# 2003 USAMO Problems/Problem 5

## Problem

Let $a$, $b$, $c$ be positive real numbers. Prove that $\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$

## Solution

### Solution 1

Since all terms are homogeneous, we may assume WLOG that $a + b + c = 3$.

Then the LHS becomes $\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \sum \frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \sum \left(\frac {1}{3} + \frac {8a + 6}{3a^2 - 6a + 9}\right)$.

Notice $3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \ge 6$, so $\frac {8a + 6}{3a^2 - 6a + 9} \le \frac {8a + 6}{6}$.

So $\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8$, as desired.

### Solution 2

Note that \begin{align*} (2x + y)^2 + 2(x - y)^2 &= 4x^2 + 4xy + y^2 + 2x^2 - 4xy + 2y^2 \\ &= 3(2x^2 + y^2). \end{align*} Setting $x = a$ and $y = b + c$ yields $$(2a + b + c)^2 + 2(a - b - c)^2 = 3(2a^2 + (b + c)^2).$$ Thus, we have $$\frac{(2a + b + c)^2}{2a^2 + (b + c)^2} = \frac{3(2a^2 + (b + c)^2) - 2(a - b - c)^2}{2a^2 + (b + c)^2} = 3 - \frac{2(a - b - c)^2}{2a^2 + (b + c)^2},$$ and its analogous forms. Thus, the desired inequality is equivalent to $$\frac{(a - b - c)^2}{2a^2 + (b + c)^2} + \frac{(b - c - a)^2}{2b^2 + (c + a)^2} + \frac{(c - a - b)^2}{2c^2 + (a + b)^2}\geq\frac{1}{2}.$$ Because $(b + c)^2\leq 2(b^2 + c^2)$, we have $2a^2 + (b + c)^2\leq 2(a^2 + b^2 + c^2)$ and its analogous forms. It suffices to show that $$\frac{(a - b - c)^2}{2(a^2 + b^2 + c^2)} + \frac{(b - c - a)^2}{2(a^2 + b^2 + c^2)} + \frac{(c - a - b)^2}{2(a^2 + b^2 + c^2)}\geq\frac{1}{2},$$ or, $$(a - b - c)^2 + (b - a - c)^2 + (c - a - b)^2\geq a^2 + b^2 + c^2.\qquad\qquad (*)$$ Multiplying this out the left-hand side of the last inequality gives $3(a^2 + b^2 + c^2) - 2(ab + bc + ca)$. Therefore the inequality $(*)$ is equivalent to $2[a^2 + b^2 + c^2 - (ab + bc + ca)]\geq 0$, which is evident because $$2[a^2 + b^2 + c^2 - (ab + bc + ca)] = (a - b)^2 + (b - c)^2 + (c - a)^2.$$ Equality holds when $a = b = c$.

### Solution 3

Given a function $f$ of three variables, define the cyclic sum $$\sum_{\text{cyc}}f(p,q,r) = f(p,q,r) + f(q,r,p) + f(r,p,q).$$ We first convert the inequality into $$\sum_{\text{cyc}}\frac{2a(a + 2b + 2c)}{2a^2 + (b + c)^2}\leq 5.$$ Splitting the 5 among the three terms yields the equivalent form $$\sum_{\text{cyc}}\frac{4a^2 - 12a(b + c) + 5(b + c)^2}{3[2a^2 + (b + c)^2]}\geq 0.\qquad\qquad (2)$$ The numerator of the term shown factors as $(2a - x)(2a - 5x)$, where $x = b + c$. We will show that $$\frac{(2a - x)(2a - 5x)}{3(2a^2 + x^2)}\geq -\frac{4(2a - x)}{3(a + x)}.\qquad\qquad (3)$$ Indeed, $(3)$ is equivalent to $$(2a - x)[(2a - 5x)(a + x) + 4(2a^2 + x^2)]\geq 0,$$ which reduces to $$(2a - x)(10a^2 - 3ax - x^2) = (2a - x)^2(5a + x)\geq 0,$$ evident. We proved that $$\frac{4a^2 - 12a(b + c) + 5(b + c)^2}{3[2a^2 + (b + c)^2]}\geq -\frac{4(2a - b - c)}{3(a + b + c)},$$ hence $(2)$ follows. Equality holds if and only if $2a = b + c, 2b = c + a, 2c = a + b$, i.e., when $a = b = c$.

### Solution 4

Given a function $f$ of $n$ variables, we define the symmetric sum $$\sum_{\text{sym}}f(x_1, \ldots, x_n) = \sum_{\sigma} f(x_{\sigma(1)}, \ldots, x_{\sigma(n)})$$ where $\sigma$ runs over all permutations of $1, \ldots, n$ (for a total of $n!$ terms).

We combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is $$\sum_{\text{sym}} 8a^6 + 8a^5 + 2a^4b^2 + 10a^4bc + 10a^3b^3 - 52a^3b^2c + 14a^2b^2c^2.$$ Recalling Schur's Inequality, we have $$a^3 + b^3 + c^3 + 3abc - (a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2) \\ = a(a - b)(a - c) + b(b - a)(b - c) + c(c - a)(c - b)\geq 0,$$ or $$\sum_{\text{sym}} a^3 - 2a^2b + abc\geq 0.$$ Hence, $$0\leq 14abc\sum_{\text{sym}} a^3 - 2a^2b + abc = \sum_{\text{sym}} 14a^4bc - 28a^3b^2c + 14a^2b^2c^2$$ and by repeated AM-GM inequality, $$0\leq\sum_{\text{sym}} 4a^6 - 4a^4bc$$ and $$0\leq\sum_{\text{sym}} 4a^6 + 8a^5b + 2a^4b^2 + 10a^3b^3 - 24a^3b^2c.$$ Adding these three inequalities yields the desired result.

## Solution 5

Since, the inequality is homogeneous we may assume that $a+b+c=1$ and $0.

The first time on the LHS is the inequality will be: $$f(a)=\frac{(a+1)^2}{2a^2+(1-a)^2}=\frac{a^2+2a+1}{3a^2-2a+1}$$ Note that equality holds when $a=b=c=1/3$. A simple sketch of $f(x)$ on $[0,1]$ shows that the curve lies below the tangent line at $1/3$.

Which has the equation of the form $y=\frac{12x+4}{3}$.

So we claim that $$f(a) = \frac{a^2+2a+1}{3a^2-2a+1} \leq \frac{12a+4}{3} \text{ for } 0 Upon clearing the denominators, it is equivalent to: $\[36a^3-15a^2-2a+1 \geq 0$$

Note that since the curve and the line intersect at $1/3, 3a-1$ would be a factor. $$36a^3-15a^2-2a+1 = (3a-1)^2(4a+1) \geq 0 \text{ for } 0 Adding the similar inequalities for $b$ and $c$ gives: $\[f(a)+f(b)+f(c) \leq \frac{12(a+b+c) +12}{3} = 8$$

## Solution 6

As in previous solutions, assume WLOG $a+b+c = 1$ and define $$f(x) = \frac{x^2+2x+1}{3x^2-2x+1}.$$ Bashing out the second derivative, it turns out that $f(x)$ is concave, so we can use Jensen's. We see that $$f(a)+f(b)+f(c) \le 3f\left(\frac{a+b+c}{3}\right) = 3f(1/3) = 8,$$ and we are done.

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