$\sum\limits_{\text{cyc}}\frac{a}{\sqrt{a^2+8bc}}\ge1$ (IMO 2001 problem 2) is mentioned in many books about inequality. However, the upper bound is seldom talked about.
Let a, b and c be positive real numbers. Prove that:
$$\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}<2$$
Or let a, b and c be non-negative real numbers and $ab+bc+ca>0$,
$$\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\le2$$
The equality holds if one of three is 0.
Firstly I don't think isolated fudging works because it's hard to find a polynomial fraction about a, b and c, e.g. $f=\frac{ka^2+mab+nb^2+\dots}{p(a^2+b^2+c^2)+q(ab+bc+ca)}$, because $f\vert_{a=0}=0$ for all b and c (we also need $f\vert_{b=0}=0$ for all c and a, etc) seems impossible.
And I think it's hard to apply AM-GM, because the equality doesn't hold when three radicals are equal.
Best Answer
Problem. Let $a, b, c$ be positive real numbers. Prove that: $$\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}<2.$$
Proof. It suffices to prove that $$\frac{a}{\sqrt{a^2+2bc}}+\frac{b}{\sqrt{b^2+2ca}}+\frac{c}{\sqrt{c^2+2ab}}<2.$$
Let $x := \frac{2bc}{a^2}, y := \frac{2ca}{b^2}, z := \frac{2ab}{c^2}$. Then $xyz = 8$. We need to prove that $$\frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + y}} + \frac{1}{\sqrt{1 + z}} < 2.$$
WLOG, assume that $x \le y \le z$.
We split into two cases:
Case 1. $x + y \ge 6$
We have $y \ge 3$. Thus, $$\mathrm{LHS} < 1 + \frac{1}{\sqrt{1 + 3}} + \frac{1}{\sqrt{1 + 3}} = 2.$$
Case 2. $x + y < 6$
Using AM-GM, we have $$\frac{1}{\sqrt{1 + x}} = \frac{2\sqrt{1 + x}}{2(1 + x)} \le \frac{(1 + x) + 1}{2(1 + x)} = \frac{x + 2}{2 + 2x}$$ and $$\frac{1}{\sqrt{1 + y}} = \frac{2\sqrt{1 + y}}{1 + y} \le \frac{(1 + y) + 1}{2(1 + y)} = \frac{y + 2}{2 + 2y}.$$
It suffices to prove that $$\frac{x + 2}{2 + 2x} + \frac{y + 2}{2 + 2y} + \frac{1}{\sqrt{1 + z}} < 2$$ or $$\frac{x}{2 + 2x} + \frac{y}{2 + 2y} > \frac{1}{\sqrt{1 + z}}.$$
Using AM-GM, it suffices to prove that $$ 2\sqrt{\frac{x}{2 + 2x} \cdot \frac{y}{2 + 2y}} > \frac{1}{\sqrt{1 + z}}$$ or $$4\cdot \frac{x}{2 + 2x} \cdot \frac{y}{2 + 2y} > \frac{1}{1 + z}$$ or (using $z = \frac{8}{xy}$) $$\frac{xy(7 - x - y)}{(1 + x)(1 + y)(xy + 8)} > 0$$ which is true.
We are done.