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 $0<a,b, c<1$.

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 \[f(a) = \frac{a^2+2a+1}{3a^2-2a+1} \leq \frac{12a+4}{3} \text{ for } 0 <a <1\]

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.

\[36a^3-15a^2-2a+1 = (3a-1)^2(4a+1) \geq 0 \text{ for } 0<a<1\]

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.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

2003 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS