Prove that $\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$
I think $uvw$ does not help here.
My another similar inequality is very easy:
with the same condition prove that:
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$
My proof: 
By AM-GM $(a+b+c)^3=a^3+b^3+c^3+24\geq9\sqrt[9]{(a^3+b^3+c^3)\cdot3^8}$
and we are done!
Thank you!
 A: Another solution using KKT conditions (or Lagrange multiplier)
Let
\begin{align}
L = \ln \frac{a+b+c}{3} - \frac{1}{53}\ln \frac{a^4+b^4+c^4}{3} 
- t ((a+b)(b+c)(c+a) - 8).
\end{align}
From $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial b} = \frac{\partial L}{\partial c} = 0$, we have
\begin{align}
0 = \frac{\partial L}{\partial a} - \frac{\partial L}{\partial b}= 
\frac{\left(a - b\right)\, \left((53a^5+53a^4b+53ab^4+53ac^4+53b^5+53bc^4)t-4a^2-4ab-4b^2\right)}{53(a^4 + b^4 + c^4)},\quad \cdots\cdots\cdots (2)\\
0 = \frac{\partial L}{\partial b} - \frac{\partial L}{\partial c}
= \frac{\left(b - c\right)\, \left((53a^4b+53a^4c+53b^5+53b^4c+53bc^4+53c^5)t-4b^2-4bc-4c^2\right)}{53(a^4 + b^4 + c^4)}. \quad \cdots\cdots\cdots(3)
\end{align}
We claim that two of $a, b, c$ are equal. Otherwise, from (2), we have
$$t= \frac{4(a^2+ab+b^2)}{53(a^5 + a^4\, b + a\, b^4 + a\, c^4 + b^5 + b\, c^4)}$$
which, when inserted into (3), results in
$$\frac{4}{53}\frac{\left(a - c\right)\, \left(b-c\right)\, \left(a\, b + a\, c + b\, c\right)}{\left(a + b\right)\, \left(a^4 + b^4 + c^4\right)} =0.$$
Contradiction.
Further we claim that $a = b = c$. Otherwise, WLOG, assume that $a = b \ne c$, from (3), we have
$$t = \frac{4}{53}\frac{a^2 + a\, c + c^2}{2\, a^5 + 2\, a^4\, c + a\, c^4 + c^5}$$
which results in
$$0 = \frac{\partial L}{\partial a} = -\frac{1}{53}\frac{ - 74\, a^4 + 48\, a^3\, c + 48\, a^2\, c^2 + 24\, a\, c^3 - 49\, c^4}{\left(2\, a^4 + c^4\right)\, \left(2\, a + c\right)}.$$
However, $- 74\, a^4 + 48\, a^3\, c + 48\, a^2\, c^2 + 24\, a\, c^3 - 49\, c^4\ne 0$
which follows from $74x^4-48x^3-48x^2-24x+49  > 0$ for all real numbers $x$. Contradiction. We are done.
