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Difference between revisions of "1997 USAMO Problems/Problem 5"

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4. Cancel like mad.
 
4. Cancel like mad.
  
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5. You are left with <math>a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3</math>. Homogenize the inequality by multiplying each term of the LHS by <math>a^2b^2c^2</math>. Because <math>(6, 3, 0)</math> ''majorizes'' <math>(5, 2, 2)</math>, this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is <math>\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2</math>.)
+
5. You are left with <math>a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3</math>. Homogenize the inequality by multiplying each term of the LHS by <math>a^2b^2c^2</math>. Because <math>(6, 3, 0)</math> ''majorizes'' <math>(5, 2, 2)</math>, this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is <math>\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2</math>. Sum similar expressions to obtain the desired result.)
  
 
==See Also ==
 
==See Also ==

Revision as of 22:05, 22 April 2014

Problem

Prove that, for all positive real numbers $a, b, c,$

$(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}$.

Prove that, for all positive real numbers $a, b, c,$

$(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}$.

Solution

USAMO97(5-solution).jpg

Solution 2

Outline:

1. Because the inequality is homogenous, scale $a, b, c$ by an arbitrary factor such that $abc = 1$.

2. Replace all $abc$ with 1. Then, multiply both sides by $(a^3 + b^3 + 1)(b^3 + c^3 + 1)(a^3 + c^3 + 1)$ to clear the denominators.

3. Expand each product of trinomials.

4. Cancel like mad.

5. You are left with $a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3$. Homogenize the inequality by multiplying each term of the LHS by $a^2b^2c^2$. Because $(6, 3, 0)$ majorizes $(5, 2, 2)$, this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is $\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2$. Sum similar expressions to obtain the desired result.)

See Also

1997 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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