# Prove that: $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{\sqrt{a^2+3bc}}+\frac{b}{\sqrt{b^2+3ac}}+\frac{c}{\sqrt{c^2+3ab}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$$ I tried Cauchy-Schwarz: $$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{a^2+3bc}.$$ Hence, it remains to prove that $$\sum_{cyc}\frac{a}{a^2+3bc}\leq\frac{81(a^2+b^2+c^2)}{4(a+b+c)^5},$$ which is wrong for $c\rightarrow0^+$.

Also we can use the following C-S: $$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\right)^2\leq(1+1+1)\sum_{cyc}\frac{a^2}{a^2+3bc}.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{a^2}{a^2+3bc}\leq\frac{27(a^2+b^2+c^2)^2}{4(a+b+c)^4},$$ which is wrong again: $b=c=1$.

• I tried to fix the missing square on the RHS numerator of your third equation, but it would not let me (the change was too small :-). Maybe you can fix that. – NickD Mar 22 '17 at 22:27

Using the substitutions $$(a, b, c) \to (a^2, b^2, c^2)$$, the inequality becomes $$\sum_{\mathrm{cyc}}\frac{a^2}{\sqrt{a^4+3b^2c^2}} \le \frac{9(a^4+b^4+c^4)}{2(a^2+b^2+c^2)^2}.$$ Using the Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align} \mathrm{LHS}^2 &= \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2} + \sum_{\mathrm{cyc}} \frac{2a^2b^2}{\sqrt{(a^4+3b^2c^2)(b^4+3c^2a^2)}}\\ &\le \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2} + \sum_{\mathrm{cyc}} \frac{2a^2b^2}{a^2b^2 + 3abc^2}. \end{align} It suffices to prove that $$\frac{81(a^4+b^4+c^4)^2}{4(a^2+b^2+c^2)^4}\ge \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2} + \sum_{\mathrm{cyc}} \frac{2a^2b^2}{a^2b^2 + 3abc^2}.$$ After clearing the denominators, it suffices to prove that $$f(a, b, c)\ge 0$$ where $$f(a,b,c)$$ is a homogeneous polynomial of degree $$26$$. Due to symmetry and homogeneity, WLOG, assume that $$1 = c \le b \le a$$. Let $$b = 1 + s, \ a = 1+s + t; \ s,t \ge 0$$. $$f(1+s+t, 1+s, 1)$$ is a polynomial in $$s, t$$ with non-negative coefficients. We are done.