1997 USAMO Problems/Problem 5

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.)

Solution 3 (Isolated fudging)

Because the inequality is homogenous (i.e. $(a, b, c)$ can be replaced with $(ka, kb, kc)$ without changing the inequality other than by a factor of $k^n$ for some $n$ ), without loss of generality, let $abc = 1$ .

Lemma: $\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.$ Proof: Rearranging gives $(a^3 + b^3) c + c \ge a + b + c$ , which is a simple consequence of $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.$

Thus, by $abc = 1$ : $\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}$ $\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.$

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

Contents

Problem

Solution

Solution 2

Solution 3 (Isolated fudging)

See Also