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

Revision as of 02:41, 18 July 2014

Problem

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

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

Solution

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.)

@@ Line 1: / Line 1: @@
 == Problem ==
-Prove that, for all positive real numbers <math>a, b, c,</math>
-<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.
 Prove that, for all positive real numbers <math>a, b, c,</math>

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

Page

Toolbox

Search

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

Revision as of 02:41, 18 July 2014

Contents

Problem

Solution

Solution 2

See Also