Inequality with $abc=2$ Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $\frac{a+b+c}{4}\geq \sqrt[4]{\frac{a^2+b^2+c^2}{6}}$
Sorry, I don't have an idea for this problem.If I had, I would have showed them all.
We need to prove that:
$$\left(\frac{a+b+c}{4}\right)^{12}\geq\left(\frac{a^2+b^2+c^2}{6}\right)^3\cdot\frac{a^2b^2c^2}{4}.$$
Now, by the Vasc's inequality
$(a+b+c)^5\geq81abc(a^2+b^2+c^2)$ 
We can obtain
$$a+b+c\geq\sqrt[5]{81abc(a^2+b^2+c^2)}$$
so it's enough to prove that
$$\left(\frac{\sqrt[5]{81abc(a^2+b^2+c^2)}}{4}\right)^{12}\geq\left(\frac{a^2+b^2+c^2}{6}\right)^3\cdot\frac{a^2b^2c^2}{4},$$
which is wrong for $abc\rightarrow0^+$. 

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality is equivalent to $f(v^2)\geq0$, where
$$f(v^2)=\frac{3}{4}u-\left(\frac{3u^2-2v^2}{2}\right)^{\frac{1}{4}}\left(\frac{w^3}{2}\right)^{\frac{1}{6}}.$$
We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.
Now, let $a=b=\sqrt[3]t$.
Thus, $c=\frac{2}{\sqrt[3]{t^2}}$ and we need to prove that
$$\frac{2\sqrt[3]t+\frac{2}{\sqrt[3]{t^2}}}{4}\geq\sqrt[4]{\frac{2\sqrt[3]{t^2}+\frac{4}{\sqrt[3]{t^4}}}{6}}$$ or 
$$\frac{t+1}{2}\geq\sqrt[3]t\sqrt[4]{\frac{t^2+2}{3}}$$ or $g(t)\geq0,$ where
$$g(t)=\ln\frac{t+1}{2}-\frac{1}{3}\ln{t}-\frac{1}{4}\ln\frac{t^2+2}{3}.$$
But $$g'(t)=\frac{1}{t+1}-\frac{1}{3t}-\frac{t}{2(t^2+2)}=\frac{(t-2)^2(t-1)}{6t(t+1)(t^2+2)},$$
which gives $t_{min}=1$ and since $g(1)=0,$ we are done!
