Prove that $\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$

Question

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$$ I tried SOS, C-S, Holder, Mixing Variables and more, but without success.

Answer 1

The Buffalo Way works although it is an ugly solution.

WLOG, assume that $a+b+c=3$. Write the inequality as $$\sum_{\mathrm{cyc}} \sqrt{\frac{a^2+10bc}{11}} \ge \sum_{\mathrm{cyc}} \sqrt{\frac{a(b+c)}{2}}.$$

Let $X = \frac{a^2+10bc}{11}, \ Y = \frac{b^2+10ca}{11}, \ Z = \frac{c^2+10ab}{11}$ and $U = \frac{a(b+c)}{2}, \ V = \frac{b(c+a)}{2}, \ W = \frac{c(a+b)}{2}$.

We will use the following bounds: $$\sqrt{x} \le \frac{(x+1)(x^2+14x+1)}{2(x+3)(3x+1)}, \quad \forall x > 0$$ and $$\sqrt{x} \ge \frac{2(x+3)(3x+1)x}{(x+1)(x^2+14x+1)}, \quad \forall x > 0$$ which follows from $$\Big(\frac{(x+1)(x^2+14x+1)}{2(x+3)(3x+1)}\Big)^2 - x = \frac{(x-1)^6}{4(x+3)^2(3x+1)^2}.$$

With the bounds above, it suffices to prove that $$\sum_{\mathrm{cyc}(X,Y,Z)} \frac{2(X+3)(3X+1)X}{(X+1)(X^2+14X+1)} \ge \sum_{\mathrm{cyc}(U,V,W)}\frac{(U+1)(U^2+14U+1)}{2(U+3)(3U+1)}.$$

After homogenization, it suffices to prove that $f(a,b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial with degree $32$.

We use the Buffalo Way. Due to symmetry, assume that $c\le b\le a$.

If $c = 0$, $f(a,b,0)$ is a polynomial with non-negative coefficients. True.

If $c > 0$, let $c=1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$. $f(1+s+t, 1+s, 1)$ is a polynomial with non-negative coefficients. True.

We are done.