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.

I repeat the above hints:
Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc.  Further, we have $a^3+b^3+c^3 = (a+b+c)^3 -24 = (x+y+z)^3 -24$.
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. 
So we have  to show
$$f = ((x+z+y)^3 -24)^2- 3 \sum_{cyc}(x+z-y)^4 \geq 0$$
From the parametrization, we see that $f$ is periodic in $q$ with period length $2 \pi/3$. The maxima are at $q_+ = n 2 \pi/3$ and the minima are at $q_- = \pi/3 + n 2 \pi/3$. Hence, it is enough to investigate $f$ at $q = \pi/3$. This gives
$$f(r,q) \ge f(r,q = \pi/3) = \\
(\exp(-3r)(2\exp(3r/2) + 1)^3 - 24)^2 - 6\exp(-4r) - 3\exp(-4r)(2\exp(3r/2) - 1)^4$$
It is now a matter of calculus to show that the minimum of $f(r,q = \pi/3)$ occurs at $r=0$, giving $f(r,q) \ge f(r=0,q = \pi/3) = 0 \geq 0$ which proves the claim. $\qquad \Box$
