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Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\frac{a}{a^2+b^2+4}+\frac{b}{b^2+c^2+4}+\frac{c}{c^2+a^2+4}\leq\frac{1}{2}$$

This inequality is a similar to the following.

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\frac{a}{a^2+b^2+2}+\frac{b}{b^2+c^2+2}+\frac{c}{c^2+a^2+2}\leq\frac{3}{4},$$ which we can prove by AM-GM: $$\sum_{cyc}\frac{a}{a^2+b^2+2}\leq\frac{1}{2}\sum_{cyc}\frac{a}{\sqrt{(a^2+1)(b^2+1)}}\leq\frac{1}{4}\sum_{cyc}\left(\frac{a^2}{a^2+1}+\frac{1}{b^2+1}\right)=\frac{3}{4}$$

Michael Rozenberg
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1 Answers1

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The Buffalo Way works.

With the substitutions $a = \frac{x^2}{yz}, \ b = \frac{y^2}{zx}, \ c = \frac{z^2}{xy}; \ x, y, z > 0$, after clearing denominators, it suffices to prove that $f(x, y, z) \ge 0$ where $f(x,y,z)$ is a homogeneous polynomial.

WLOG, assume that $z = \min(x, y, z) = 1$. There are two possible cases:

1) $1 = z \le x \le y$: Let $x = 1+s, \ y = 1+s + t; \ s, t\ge 0$. Then $f(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true.

2) $1 = z \le y\le x$: Let $y = 1+s, \ x = 1+s+t; \ s,t\ge 0$. Then \begin{align} f(1+s+t, 1+s, 1) &= (4s^2t^{14} - 2s^4t^{13} + \frac{1}{4}s^6t^{12}) + (\frac{3}{4}s^6 - 20s^5+ 249s^4)t^{12} \\ &\quad + (12s^7 - 72s^6 + 772s^5)t^{11} + (66s^8-46s^7+1834s^6)t^{10} + g(s,t) \end{align} where $g(s,t)$ is a polynomial with non-negative coefficients. Clearly, $f(1+s+t, 1+s, 1)\ge 0$. We are done.

River Li
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