Inequality $\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$
Some attempts:


*

*From the condition follows $a^3+b^3+c^3 = (a+b+c)^3 -24$

*It is known (see here) 
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$

*Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc.  The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x =  \exp(r \cos q) \; ; \;  y =  \exp(r \cos (q + 2 \pi /3)) \; ; \;  z =  \exp(r \cos (q - 2 \pi /3)) 
$$
Using that, the condition can be removed and then calculus may be used. 

*The question may be interpreted geometrically. Expressions such as $a^3+b^3+c^3 = $const. and $a^4+b^4+c^4 =$ const. can be  interpreted as hypersurfaces of what has been called  an N(3)-dimensional ball in p-norm, see here. A nice visualization is given in here. Then properties such as extrema, convexity etc. of these surfaces can be used.  
I couldn't put the pieces together.

We need to prove that $$\left(\frac{a^3+b^3+c^3}{3}\right)^6\geq\left(\frac{a^4+b^4+c^4}{3}\right)^3\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^2$$ or
$$\left(\frac{a^3+b^3+c^3}{3}\right)^6\geq\left(\frac{a^4+b^4+c^4}{3}\right)^3\left(\frac{(a+b+c)^3-a^3-b^3-c^3}{24}\right)^2.$$
Now, let $a+b+c$ be constant and $a^3+b^3+c^3$ be constant.
Thus, by the Vasc's EV Method:
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
Corollary 1.8, case 3(a)
$a^4+b^4+c^4$ gets a maximal value for an equality case of two variables.
Since the last inequality is homogeneous, 
it's enough to check what happens for $b=c=1$, which gives
$$\left(\frac{a^3+2}{3}\right)^6\geq\left(\frac{a^4+2}{3}\right)^3\left(\frac{(a+1)^2}{4}\right)^2$$ or $f(a)\geq0,$ where
$$f(a)=6\ln(a^3+2)-3\ln(a^4+2)-4\ln(a+1)-3\ln3+4\ln2.$$
Now, $$f'(a)=\frac{2(a-1)(a^6+4a^5+4a^4-12a^3-10a^2+8a+8)}{(a^3+2)(a^4+2)(a+1)}$$
and since $$a^6+4a^5+4a^4-12a^3-10a^2+8a+8>0,$$ we are done!
By the same way we can prove your stronger inequality:
$$\sqrt[27]{\frac{a^3+b^3+c^3}{3}}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$ and my old
$$\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$ is proven!
