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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.

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    Michael! Meeting you again. This time too, I'm without a proof. But the extremum is achieved when $a = b = c = 1$. :-) – 2017-02-07
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    @Nilabro Saha This inequality is cyclic and not symmetric, but we see that it indeed happens. – 2017-02-07
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    Yea, true. This is another proof that my reasoning was actually flawed. – 2017-02-07

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