If $ab+ac+bc=3$ so $\sum\limits_{cyc}\sqrt{a^2+b^2+7bc}\geq9$.

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$. Prove that: $$\sqrt{a^2+b^2+7bc}+\sqrt{b^2+c^2+7ca}+\sqrt{c^2+a^2+7ab}\geq9$$ The equality occurs also for $a=2$, $b=\frac{3}{2}$ and $c=0$.

I tried the following Holder. $$\left(\sum\limits_{cyc}\sqrt{a^2+b^2+7bc}\right)^2\sum_{cyc}(a^2+b^2+7bc)^2(ka+mb+c)^3\geq$$ $$\geq\left(\sum_{cyc}(a^2+b^2+7bc)(ka+mb+c)\right)^3.$$ Thus, it remains to prove that $$\left(\sum_{cyc}(a^2+b^2+7bc)(ka+mb+c)\right)^3\geq9(ab+ac+bc)^2\sum_{cyc}(a^2+b^2+7bc)^2(ka+mb+c)^3,$$ but I did not find a non-negative values of $k$ and $m$, for which the last inequality would be true.

• Michael! Meeting you again. This time too, I'm without a proof. But the extremum is achieved when $a = b = c = 1$. :-) – Nilabro Saha Feb 7 '17 at 20:07
• @Nilabro Saha This inequality is cyclic and not symmetric, but we see that it indeed happens. – Michael Rozenberg Feb 7 '17 at 20:10
• Yea, true. This is another proof that my reasoning was actually flawed. – Nilabro Saha Feb 7 '17 at 20:12

By Holder, we have \begin{align} &\Big(\sum_{\mathrm{cyc}} \sqrt{a^2+b^2+7bc}\Big)^2\sum_{\mathrm{cyc}} (a^2+b^2+7bc)^2 (6ab + 3bc + 10ca)^3\\ \ge \ & \Big(\sum_{\mathrm{cyc}} (a^2+b^2+7bc)(6ab+3bc+10ca)\Big)^3. \end{align} Thus, it suffices to prove that \begin{align} &\Big(\sum_{\mathrm{cyc}} (a^2+b^2+7bc)(6ab+3bc+10ca)\Big)^3\\ \ge \ & 27(ab+bc+ca)\sum_{\mathrm{cyc}} (a^2+b^2+7bc)^2 (6ab + 3bc + 10ca)^3. \end{align} Let $$f(a,b,c) = \mathrm{LHS}-\mathrm{RHS}$$.
The Buffalo Way works. WLOG, assume that $$c = \min(a,b,c)$$. There are two possible cases:
1) $$c\le b \le a$$: Let $$b = c + s,\ a = c+s+t; \ s,t\ge 0$$. Then $$f(c+s+t, c+s, c) = g(c,s,t) + s^3(352s^4+1068s^3t+1269s^2t^2+378st^3+81t^4)(s-3t)^2(s+t)^3$$ where $$g(c,s,t)$$ is a polynomial with non-negative coefficients. True.
2) $$c\le a\le b$$: Let $$a=c+s, \ b = c+s+t; \ s,t\ge 0$$. Then $$f(c+s, c+s+t, c)$$ is a polynomial in $$c, s, t$$ with non-negative coefficients. True.