TJUSAMO 2013-2014 Inequalities

TJUSAMO 2013-2014Inequalities Robin Park Inequalities Robin Park January 26, 2014 1 Definitions Definition 1.1. A function is said to be convex (or concave up) if the line segment between any two points on the graph of the function lies above the graph. A function is said to be concave (or concave down) if the line segment between any two points on the graph of the function lies below the graph. Formally, a function f : I → R is concave up if for any two points x1 , x2 ∈ I and t ∈ [0, 1], f (tx1 + (1 − t)x2 ) ≤ tf (x1 ) + (1 − t)f (x2 ). The sign of the inequality is flipped if f is concave down. An easy way to check if f is convex in an interval is by taking its second derivative, and checking that it is nonnegative for the interval. Definition 1.2. A sequence a1 , a2 , · · · , an is said to majorize the sequence b1 , b2 , · · · , bn if k X ai ≥ i=1 k X bi i=1 for all k = 1, 2, · · · , n, and is written as (a1 , a2 , · · · , an )  (b1 , b2 , · · · , bn ). Definition 1.3. A shorthand way of writing inequalities is by using summation notation. Define the cyclic sum of f (a, b, c) by X f (a, b, c) = f (a, b, c) + f (b, c, a) + f (c, a, b) cyc and define the symmetric sum of f (a, b, c) by X f (a, b, c) = f (a, b, c) + f (a, c, b) + f (b, a, c) + f (b, c, a) + f (c, a, b) + f (c, b, a). sym Cyclic products and symmetric products are defined analogously. Definition 1.4. The gradient of a function f is defined as ∇f = 2 ∂f ∂f ∂f ˆ ˆı + ˆ + k. ∂x ∂y ∂z The Standard Dozen As listed in the beginning of Thomas Mildorf’s Olympiad Inequalities 1 , the following twelve inequalities are the most famous in Olympiad mathematics. 1 http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf 1 Let a. b. Let x1 .2 (Weighted Power Mean). Then a1 + a2 + · · · + an b1 + b2 + · · · + bn a1 bn + a2 bn−1 + · · · + an b1 a1 b1 + a2 b2 + · · · + an bn ≥ ≥ . we have a1 b1 + a2 b2 + · · · + an bn ≥ a1 bπ(1) + a2 bπ(2) + · · · + an bπ(n) ≥ a1 bn + a2 bn−1 + · · · + an b1 .3 (H¨ older).TJUSAMO 2013-2014 Inequalities Robin Park Theorem 2. · · · . an . Then f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ f (y1 ) + f (y2 ) + · · · + f (yn ). c be nonnegative reals and r > 0. xn majorizes the sequence y1 . Then for any x1 . and let λa . Then p p p d1 ≥ d2 ≥ 3 d3 ≥ · · · ≥ n dn . · · · . z2 . n}. y2 . then  f (r) = ω1 xr1 + ω2 xr2 + · · · + ωn xrn ω1 + ω2 + · · · + ωn  r1 is a non-decreasing function of r. Theorem 2. 0].10 (Popoviciu). ω1 + ω2 + · · · + ωn Theorem 2. b2 .   ω1 x1 + ω2 x2 + · · · + ωn xn ω1 f (x1 ) + ω2 f (x2 ) + · · · + ωn f (xn ) ≥ (ω1 + ω2 + · · · + ωn )f . Theorem 2.9 (Majorization). x2 . x2 . λb . xn ∈ I and any nonnegative reals ω1 . · · · . · · · . If x1 . a2 . Let f : I → R be a convex function on I and suppose that the sequence x1 . Then (i) d2i ≥ di+1 di−1 . yi ∈ I. Theorem 2. yn . c are equal and the other is 0. · · · . Let f : I → R be a convex function. x2 . Then for any 2 . · · · . xn be nonnegative real numbers. Theorem 2. Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Let a1 . · · · .1 (Jensen). ω2 . · · · . z1 . x2 .5 (Chebyshev). · · · . Let di be defined as above. In particular. b1 .4 (Rearrangement). Then. where if some xi is zero f is identically 0. f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality.8 (Maclaurin). Theorem 2. f is strictly increasing unless all the xi are equal except possible for r ∈ (−ω. zn be sequences of nonnegative real numbers. bn . · · · . with the convention that r = 0 is the weighted gometric mean. Let f : I → R be a convex function on I and let x. · · · . z ∈ I. · · · . 2. sn by (x + x1 )(x + x2 ) · · · (x + xn ) = sn xn + · · · + s1 x + s0 . n n n n Theorem 2. λz be positive reals which sum to 1. where xi . s1 .7 (Newton). Theorem 2. · · · . Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. and define the symmetric averages by di = sni . for any permutation π of {1. xn are nonnegative reals with a positive sum. Define the symmetric polynomials s0 . Then ar (a − b)(a − c) + br (b − c)(b − a) + cr (c − a)(c − b) ≥ 0 with equality if and only if a = b = c or some two of a.6 (Schur). ωn . Then (a1 + a2 + · · · + an )λa (b1 + b2 + · · · + bn )λb · · · (z1 + z2 + · · · + zn )λz ≥ aλ1 a bλ1 b · · · z1λz + · · · + aλna bλnb · · · znλz . y. b. Theorem 2. Then for any positive reals x1 . bn .11 (Bernoulli). Let x. Let x. where I = [a. xa1 1 xa2 2 · · · xann ≥ sym sym 3 Monotonicity Monotonicity of functions can be used to prove that the left side increases faster than the right side. 3
7 implying that f is increasing on [0. Prove the inequality 0 ≤ xy + yz + zx − 2xyz ≤ 7 . 31 ]. and r by p = x + y + z. Suppose the sequence a1 . Thus Now we prove the right inequality. q = xy +Pyz + zx. x2 . Example 3. xn . so that the left side is always greater than the right side. Let f : I → R and g : I → R be functions. and z be positive reals and define the variables p. where and are all taken to be cyclic. and so xy + yz + zx − 2xyz = (1 − 3x)yz + xy + zx + xyz ≥ 0. Q r = xyz. 4 1−x 2 2 (1 − 2x) = −2x3 + x2 + 1 . WLOG let 0 ≤ x ≤ y ≤ z ≤ 1. For all r ≥ 1 and x ≥ −1. p+q q+r r+p Theorem 2. r. Hence x ≤ 13 . b].   px + qy + rz pf (x) + qf (y) + rf (z) + (p + q + r)f p+q+r       px + qy qy + rz rz + px ≥ (p + q)f + (q + r)f + (r + p)f .12 (Muirhead). 0 4  pqr Method The pqr method is a method of solving inequalities by making substitutions and simplifying the inequality. as desired. yz ≤ y+z 2 2  xy + yz + zx − 2xyz = x(y + z) + yz(1 − 2x) ≤ x(1 − x) + Let f (x) = −2x3 +x2 +1 . · · · . 27 Solution. which is the left inequality. By AM-GM. y.1. · · · . f (x) ≤ f 31 = 27 . y. and z be nonnegative real numbers such that x + y + z = 1. a2 . 3 . q.TJUSAMO 2013-2014 Inequalities Robin Park positive reals p. · · · . f and g are differentiable on I. (1 + x)r ≥ 1 + xr. such that f (a) = g(a). we see that −3x2 + x 3x f (x) = = 2 2  1 − x ≥ 0.
2
2 = 1−x . A list of relationships between the variables x.1. and f 0 (x) > g 0 (x) for all x ∈ I. Theorem 3. r are given below. q. Therefore. q. Then f (x) > g(x) for all x ∈ I. X X xb11 xb22 · · · xbnn . y. 4 Differentiating f . and p. z. an majorizes the sequence b1 . Theorem 2. b2 . x = (p2 − 2q)2 − 2(q 2 − 2pr) Q 5. (x + y)2 (y + z) = (p2 + q)2 − 4p(pq − r) P 8. (1 + x)2 (1 + y)2 = (3 + 2p + q)2 − 2(3 + p)(1 + p + q + r) P 2 12. xy(x + y) = pq − 3r Q 9. p2 q + 3pr ≥ 4q 2 12. 2p3 + 9r ≥ 7pq 9. q 3 + 9r2 ≥ 4pqr 13. p4 + 3q 2 ≥ 4p2 q 10. p4 − 5p2 q + 4q 2 + 6pr ≥ 0 3. q 3 ≥ 27r2 7. (1 + x3 ) = p3 + q 3 + r3 − 3pqr − 3pq − 3r2 + 3r + 1 There are also inequalities that relate p. x = p2 − 2q P 3 2. x y = q 3 − 3pqr − 3r2 P 14. 2p3 + 9r2 ≥ 7pqr 11. (1 + x)(1 + y) = 3 + 2p + q P 11. x = p(p2 − 3q) + 3r P 2 2 3. p3 ≥ 27r 6. p2 ≥ 3q 5. q. (1 + x) = 1 + p + q + r P 10. x (y + z) = pq − 3r P 3 3 13. pq ≥ 9r 4. (1 + x2 ) = p2 + q 2 + r2 − 2pr − 2q + 1 Q 16. p3 − 4pq + 9r ≥ 0 2. xy(x2 + y 2 ) = p2 q − 2q 2 − pr Q 15. and r to themselves: 1. x y = q 2 − 2pr P 4 4. (x + y) = pq − r P 6. pq 2 ≥ 2p2 r + 3qr 4 Robin Park . (x + y)(y + z) = p2 + q P 7. q 2 ≥ 3pr 8.TJUSAMO 2013-2014 Inequalities P 2 1. 1. The transformation from the given inequality into pqr notation is probably the trickiest part. 9(p + 1) ⇐⇒ 27(p+1)4 ≥ 256p3 . Prove that s !  r r 5/8 1 + x2 1 + y2 1 + z2 1 x+y+z + + ≤ . Let p = x+y+z = 1. 5
p 5/4 3 ⇐⇒ 4p2 3(p+1) . y. c ≥ 0. y. v 2 . The left side expands to 1 + p + q + r.com/uvwmethod 5 . that is. If a. q = xy+yz +zx. p3 ≥ 27r. 2 + q ≥ 63r ⇐⇒ 2 ≥ 54r. so r ≤ 27 . Theorem 5. w3 ≥ 0. (David Stoner) The given condition implies that p + 1 = q + r. q ≥ 9r. s 1 X 1 + x2 ≤ 3 cyc 1 + x s 4p2 . The theorems and proofs of these theorems are outlined in Tejs Knudsen’s article2 on this method. Positive real numbers x. and the rest is generally just substituting the above inequalities/identities. which is true. so 2 4p Now 9(p+1) ≤ p 4 3 4 3p . b. Example 4. 2 http://tinyurl. Now. and w3 = abc. Prove the inequality     1 1 1 1+ 1+ 1+ ≥ 64. and the right side is 64r. x y z Solution. z > 0 such that x + y + z = 1. It remains to prove 2 + q ≥ 63r. 1+x2 cyc 1+x ≤ 1X 3 cyc r 1 + x2 ≤ 1+x 4 4 p3 94 (p+1)4 ≤ 1 35 P By Jensen’s Inequality. Note that P X1−x X 1 + x2 cyc (x − 1)(y + 1)(z + 1) =p+ =p+ 1+x 1+x p+q+r+1 cyc cyc −q − 3r + p + 3 2p + 2 1−r =p+ p+1 p2 + p − r + 1 p2 + q = = . z satisfy xyz + xy + yz + zx = x + y + z + 1. 3v 2 = ab + bc + ca. The desired inequality is (x+1)(y+1)(z +1) ≥ 64xyz. In addition. b. but usually hidden. and r = xyz. p+1 p+1 =p+ We have that 3q ≤ p2 . then u ≥ v ≥ w.2 (TSTST 2012). 3 1+x 1+y 1+z 3 Solution. Theorem 5. Example 4.TJUSAMO 2013-2014 Inequalities Robin Park The implementation of the pqr method is simple. which is true since 3p+3 = p+p+p+3 ≥ uvw Method The uvw method is similar to the pqr method in that it also involves substitution.2 (Positivity). a. Let x. c ≥ 0 if and only if u. but instead we use the substitutions 3u = a + b + c.1 (Idiot). We 1 have that pq ≥ 9r. All symmetric three-variable polynomials of degree less than or equal to 5 achieve their maximum and minimum values on R if and only if (a − b)(b − c)(c − a) = 0. and on R+ iff (a − b)(b − c)(c − a) = 0 or abc = 0. Since ab(a2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a2 ) = (a − b)(b − c)(a − c)(a + b + c). Corollary 6. b. This theorem produces a few corollaries: Corollary 6. Hence. the minimum value of M is attained at the maximum of f (t) = 12t(t−1)3 (3t−2)4 . 6 34 29 > 4 27 . a + b + c) is monotonic/convex/concave. and on R+ when (a − b)(b − c)(c − a) = 0 or abc = 0. so the ABC Method The Abstract Conconcreteness Method. the given inequality is equivalent to (a − b)2 (b − c)2 (c − a)2 (a + b + c)2 ≤ M 2 (a2 + b2 + c2 )4 . Then f achieves its maximum and minimum values on R iff (a − b)(b − c)(c − a) = 0. w3 are real if and only if u2 ≥ v 2 and i h p p w3 ∈ 3uv 2 − 2u3 − 2 (u2 − v 2 )3 . Since −(w3 −(3uv 2 −2u3 ))2 ≤ 0. and c. and on R+ if and only if (a − b)(b − c)(c − a) = 0 or abc = 0.1. ab + bc + ca. 3uv 2 − 2u3 + 2 (u2 − v 2 )3 . t→∞ (3t − 2)4 81 27 lim f (t) = lim t→∞ If t is negative.TJUSAMO 2013-2014 Inequalities Robin Park Theorem 5. solving f 0 (t) = 0 shows that t = −2 yields a maximum. Let f (abc. then f achieves its maximum and minimum values on R when (a − b)(b − c)(c − a) = 0. we have 12t(t − 1)3 ≤ M 2 (3t − 2)4 . which is positive for t ∈ [1.3 (uvw). Solution. 2 (t+2) . Making the substitutions 3u = a + b + c. and w3 = abc. and then observing where the minimum and maximum values occur. implyping that the minimum value of M is 169√2 . v 2 . the inequality becomes 9u2 · (33 (−(w3 − (3uv 2 − 2u3 ))2 + (4u2 − v 2 )3 )) ≤ M 2 (9u2 − 6v 2 )4 . Dividing through by v 8 and making the substitution t = uv . and a + b + c. ab + bc + ca. ab + bc + ca. ∞). ∞). we see that f 0 (t) = 12(t−1) (3t−2)5 [1.2. abbreviated the ABC method. (IMO 2006/3) Determine the least real number M such that the inequality |ab(a2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a2 )| ≤ M (a2 + b2 + c2 )2 holds for all real numbers a.1 (ABC Theorem). Example 5. 3v 2 = ab + bc + ca. Theorem 6. 6 . the maximum of f is 12t(t − 1)3 12 4 = = . so in this interval.3. The main idea of the method is to express an inequality as a function of abc. it suffices to prove that 12u2 (u2 −v 2 )3 ≤ M 2 (3u2 −2v 2 )4 . and checking the inequality for the specific case. Thus f is increasing in Differentiating f . If the function f (abc. is an extremely powerful but limited way of proving polynomial inequalities. Now f (−2) = 4 minimum of M 2 is 329 . u. a + b + c) be a linear function/quadratic trinomial with variable abc. All symmetric three-variable polnomials of degree less than or equal to 8 with nonnegative coefficient of (abc)2 in the form f (abc. we have to prove that 2 abc(a2 + b2 + c2 ) + (a3 + b3 + c3 )(a2 + b2 + c2 ) − (a3 + b3 + c3 )(ab + bc + ca) ≥ 0. and a + b + c. (a − b)(b − c). This is the technique that the Vietnamese are wellknown for on the Inequalities forum on AoPS. For the first case. and (c − a). and so this function has degree 3 in abc. If abc = 0. or (c − a)(a − b). Sb . c > 0 be real numbers. This exact procedure is reiterated in David Arthur’s article3 on Sum of Squares for the Canada IMO Training Camp. b. Let a. 7 Sum of Squares (SOS) Sum of Squares. (c − a)2 . 2. Example 6. 5. is a brute-force but powerful method of proving 3-variable homogenous inequalities one of whose equality cases is a = b = c. 4. (b − c). a+b+c) achieve their maximum on R iff (a−b)(b−c)(c−a) = 0. and ensure the coefficients vanish when a = b = c. 3 http://tinyurl. + ≥ 2 ⇐⇒ 3 2 +c 3 2a + c 3(2a2 + b2 )(2a3 + b3 ) 2a3 which is true.4. (b − c)(c − a).2. and c. By Corollary 6. Replace (a − b)(b − c) with 21 ((a − c)2 − (a − b)2 − (b − c)2 ). 3 Both a2 + b2 + c2 and a3 + b3 + c3 can be written as polynomial expressions of abc. ab + bc + ca. The steps to expressing an inequality X into SOS form are: 1. (b − c). Group everything by (a − b). ab+bc+ca. Clearing denominators and bringing all of the terms to one side. Write everything as a multiple of (a − b). abbreviated SOS. then WLOG let a = 0. and is much better explained because the author provides an example to go along with it. Group X into one or more terms that are 0 when a = b = c. WLOG let a = b. The inquality becomes 2 bc ≥ 2 ⇐⇒ b2 + c2 + 3(b − c)2 ≥ 0. and on R+ iff (a − b)(b − c)(c − a) = 0 or abc = 0. b. or (c − a).1. Write everything as a multiple of (a − b)2 . 3. (b − c)2 . 3 b + c2 which is true. We have to prove that a2 + 2ac 2 2(a − b)4 (a + b) a2 c ≥ 0. and Sc are functions of a. The main idea of the method is to transform an inequality into the form Sa (a − b)2 + Sb (b − c)2 + Sc (c − a)2 ≥ 0 where Sa .com/davidarthursos 7 . a3 + b3 + c3 3 a + b2 + c 2 Solution. Prove that 2 ab + bc + ca abc + ≥ 2 .TJUSAMO 2013-2014 Inequalities Robin Park Corollary 6. the minimum of the left hand side occurs when (a − b)(b − c)(c − a) = 0 or abc = 0. 2a2 + 2b2 + 3c2 ). applying Majorization Inequality gives us the desired inequality. (x + 3y + 3z)(3x + 2y + 2z)(y + 3x + 3z)(3y + 2x + 2z) cyc Solution 2. It suffices to show that cyc x+3y+3z ≥ cyc 3x+2y+2z after using AM-GM and homogenizing. we have that 2 x = ta − 71 2 . b. it is possible to write any inequality in SOS form using this procedure. Example 7. cyc implying the desired inequality.TJUSAMO 2013-2014 Inequalities Robin Park Assuming that it has equality case a = b = c. and z = tc P cyc x3 y 3 z ≥ P cyc x3 y 2 z 2 . it suffices to show that 1 cyc a2 +3b2 +3c2 P 1 cyc 3a2 +2b2 +2c2 . y = b2 . Steps 4 and 5 are optional. As above.1. cyc cyc which implies that Z 1 0 X 2 t3a +3b2 +c2 −1 Z dt ≥ 0 cyc 1 X 2 t3a +2b2 +2c2 −1 dt. For proofs to the following theorems and more example problems. Since f (x) = 1 x is concave up for x > 0. a2 + 3b2 + 3c2 )  (3a2 + 2b2 + 2c2 . P P 1 1 Solution 1. P ≥ Note that (a2 + 3b2 + c2 . 3a2 + b2 + 3c2 . 2a2 + 3b2 + 2c2 . Making the substitution − 71 gives us X 2 2 2 X 2 2 2 t3a +3b +c −1 ≥ t3a +2b +2c −1 . and c satisfying a2 + b2 + c2 = 1. Let x = a2 . Solution 3. is a technique based on averaging. or just the Mixing Variables Method. Pham Kim Hung’s 8 . y = tb − 17 2 . Prove that 1 1 1 1 1 1 + + + + ≤ a2 + 2 b2 + 2 c2 + 2 6ab + c2 6bc + a2 6ca + b2 for all positive real numbers a. and z = c2 . as there aren’t many inequalities that have that extraneous (a − b)(b − c) term. Now. By Muirhead’s Inequality. abbreviated SMV.  X X 1 (x − y) + (x − z) 1 − = x + 3y + 3z 3x + 2y + 2z (x + 3y + 3z)(3x + 2y + 2z) cyc cyc   X 1 1 = (x − y) − (x + 3y + 3z)(3x + 2y + 2z) (y + 3x + 3z)(3y + 2x + 2z) cyc   X 3(x2 − y 2 ) + xz − yz = (x − y) (x + 3y + 3z)(3x + 2y + 2z)(y + 3x + 3z)(3y + 2x + 2z) cyc   X 3x + 3y + z = (x − y)2 ≥ 0. 8 SMV Method The Strongly Mixing Variables Method. Lemma 8. b2 . xn tends to the same n . Written in mathematical terminology. it remains to prove that f 3 3 . Theorem 8. a2 . x2 . 3 ≥ 2 .   b+d b+d f (a. · · · .org/wp-content/uploads/reflections/2006_6/2006_6_mixing. Let (x1 .pdf 9 . · · · . b. b. Define a transformation as follows: 1. It can be the geometric q √ a2 +b2 mean ab. xn ) ≥ f (x. · · · . Note that a + c ≤ (a + b + c + d) = . bn ) is obtained from the sequence n (a1 . d) ≤ f a. x2 . · · · . Note that       a+b a+b a b c a+b a+b c .awesomemath. each term of sequence x1 . . and so a + c ≥ ≥ 8ac > 27 2 2 a+c 27 ac. an ) ≥ f (b1 . WLOG
let a ≤ b ≤ c ≤ d. c. Hence. · · · . 3 . · · · . The main idea of the technique is proving that substituting the average of two variables in place of the two variables increases/decreases a function. Define f (a. c. 2 2 4 https://www. xn }. d) = abc + bcd + cda + dab − 176 27 abcd = ac(b + d) + 1 1 4ac 176 bd a + c − 176 ac . c . an ) by some ∆-transformation. · · · . . xn ) be a sequence of real numbers. c) ≥ f a+b . x2 . x.c = + + − + + 2 2 b+c c+a a+b a + b + 2c a + b + 2c a + b b 2(a + b) a + − = b + c c + a a + b + 2c a3 + a2 c + b2 c + b3 − 2abc − ab2 − a2 b = ≥ 0. a2 . Let us call such transformations ∆-transformations. Replace xi and xj with xi +xj 2 . and so applying the SMV Theorem.2 (SMV Theorem). b. n Example 8. Let a. and c be positive real numbers. Let f : I ⊂ Rn → R be a symmetric and continuous function such that f (a1 . After infinitely many of the above transformation. · · · . b → a+b 2 . x). 2 f (a. (b + c)(a + c)(a + b + 2c)

a+b a+b+c a+b+c a+b+c Hence f (a. Then f (x1 . and so on. · · · . Let a. and Selected Problems are good resources.1. b2 . . Techniques. implying that 176 a + c − 27 ac > 0. b+c c+a a+b 2 Solution. Denote f (a. 2. b. and then after applying this substitution infinitely many times. c) − f Example 8. 27 27 Solution. xn } and xj = min{x1 .Theorems. where x = x1 +x2 +···+x . c. b. where the sequence (b1 . Prove that b c 3 a + + ≥ . bn ). 2 . Prove the inequality 1 176 abc + bcd + cda + dab ≤ + abcd. limit x = x1 +x2 +···+x n In fact. which is clearly true.TJUSAMO 2013-2014 Inequalities Robin Park article on the SMV Method4 and Zdravko Cvetkovski’s book Inequalities .1 (Nesbitt). · · · . this transformation does not necessarily have to be the average a. b. Let xi = min{x1 . x2 . and d be four nonnegative real numbers satisfying a + b + c + d = 1.2 (ISL 1993). c) = a b+c + b c+a + c a+b . x2 . all of the variables must end up being equal. c. b. or the quadratic mean 2 . This method is highly looked down upon by Olympiad graders. 32 Solution. t). we have that f (a. and thus it is often referred to as “Lagrange murderpliers” by MOPpers. To find the maximum and minimum values of f (x. Example 10. As ∇f = h3a2 + 4bc. z). (Thomas Mildorf) Let f (a. as it involves Multivariable Calculus. b. z) subject to the constraint g(x. t. the smallest is the minimum value of f . 3b2 + 4ca. b. y. z) that result from the previous step. Moreover. z) = k (assuming that these extreme values exist): 1. z) = k. Because f and g are polynomials. c ≤ 1 2 be real numbers with a + b + c = 1. by the theorem of Lagrange multipliers. we obtain the polynomial inequality 27 (1 − 3t)(4t − 1)2 (11t + 1) ≥ 0. z) = λ∇g(x. 3c2 + 4abi 5 http://www. 10 Lagrange Multipliers The method of Lagrange multipliers is a technique for finding the maximum and minimum of a function subject to various constraints. y. t. in order to receive full points for a Lagrange multipliers solution. The largest of these values is the maximum value of f . many times and adding everything up to obtain the desired inequality. Hence. g(x. y. 9 Dumbassing To dumbass an inequality is to mindlessly brute-force the inequality by first expanding it out into a polynomial expression. the gradient of g is never zero. Evaluate f at all the points (x. b. c. where t = b+c+d 3 1 3 t2 (3a + t) ≤ 27 + 176 at given that a + 3t = 1.1. they have continuous first partial derivatives. Now we must prove that By the SMV Theorem. c) = a3 + b3 + c3 + 4abc and g(a. which is true. y. y.pdf 10 . y. such as compactification. Letting a = 1 − 3t. any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. This “art of dumbassing” is enunciated in Brian Hamrick’s article5 on Chinese Dumbass Notation. Find all values of x.TJUSAMO 2013-2014 Inequalities Robin Park . z. and then using Muirhead and AM-GM many. The method is outlined below: Theorem 10. d) ≤ f (a. c) = a + b + c − 1. y.edu/ ~2010bhamrick/files/dumbassing. and λ such that ∇f (x. Show that a3 + b3 + c3 + 4abc ≤ 9 . 2.1. b.tjhsst. Let 0 ≤ a. one must fully rigorize all of one’s steps. In addition. (IMO 2008) If x. Since h(t) has a continuous derivative. and z are real numbers. 3a2 + 4ac = 3c2 + 4a2 and a2 − 4ac + 3c2 = (a − c)(a − 3c) = 0. = 3b2 + 4ca. 1 >. y. b. 2 − t) ≤ 32 for 0 ≤ t ≤ 21 . b. and c. Let x > 0 be a real number. Hence. t. The interior local extrema therefore occur when a = b = c or when two of {a. (a + b − c)2 (b + c − a)2 (c + a − b)2 3 + + ≥ . a = b.   1 1 1 9 (ab + bc + ca) + + ≥ . We have 3a2 + 4bc = 3b2 + 4ca or (a − b)(3a + 3b − 4c) = (a − b)(3 − 7c) = 0 for any permutation of a. we could observe that h00 (t) = −1 < 0. which guarantees that h( 14 ) is a local minimum. = 3c2 + 4ab. and c. 2 2 2 2 (a + b) + c (b + c) + a (c + a)2 + b2 5 5. 1. Its derivative h0 (t) = 14 − t is zero only at t = 14 . b. 7 . without loss of generality. 13 ) = 27 49 = f ( 7 . c) = a + b + c − 1. c) is symmetric in 1 1 9 a. g(a. (As a further check. (Japan 1997) Show that for all positive reals a. 4 4 2 h(t) is 14 at either endpoint. Recalling that f (a. b. Now. 2a2 + (b + c)2 2b2 + (c + a)2 2c2 + (a + b)2 3. Prove that x − x2 2 < ln(x + 1). 7 ). 4. 41 ) = 9 32 . such that xyz = 1. (a + b)2 (b + c)2 (c + a)2 4 2. (USAMO 2003) Let a. 2 2 (x − 1) (y − 1) (z − 1)2 11 .) 11 Problems 1. the only boundary check we need is f ( 2 . and c be positive real numbers. we have λ = 3a2 + 4bc. Prove that (2a + b + c)2 (2b + c + a)2 (2c + a + b)2 + + ≤ 8. all different from 1.TJUSAMO 2013-2014 Inequalities Robin Park and ∇g =< 1. We solve   1 1 h(t) = f . t. we are done. b. Checking. b. b. (Iran 1996) Prove that for all positive real numbers a. and c. 14 . then prove that x2 y2 z2 + + ≥ 1. we have f ( 31 . c} are three times as large 7 1 3 3 < 13 as the third. − t 2 2  3   1 1 1 − t + 2t −t = + t3 + 8 2 2 2 1 t t = + − . h( 41 ) = f ( 12 . and c. b. 31 . Checking. b. Prove that r r r  √ √ √ √ a2 + b2 b2 + c2 c2 + a2 a+b+ b+c+ c+a + + +3≤ 2 a+b b+c c+a 9. π2 ).TJUSAMO 2013-2014 Inequalities Robin Park 6. and c be positive real numbers such that 1 a + 1 b + 1 c = a + b + c. 10. (ISL 2009/A4) Let a. Let a. 2b2 + 2c2 − a2 2c2 + 2a2 − b2 2a2 + 2b2 − c2 7. (Gabriel Dospinescu) Prove that for any positive real numbers a. (ISL 2009/A2) Let a. 3a 3b 3c 2a + b 2b + c 2c + a a + 2b b + 2c c + 2a 12 . and c be positive real numbers such that ab + bc + ca ≤ 3abc. (IMC 2006) Compare tan(sin x) and sin(tan x) for all x ∈ (0. b. b. b. and c the following inequality holds:     1 1 1 1 1 1 1 1 1 + + +2 + + ≥3 + + . Prove that 1 1 3 1 + + ≤ . Prove that √ √ a b c +√ +√ ≥ 3. (2a + b + c)2 (a + 2b + c)2 (a + b + 2c)2 16 8. and c be the lengths of the sides of a triangle.