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 start with two identities :
$$\frac{1}{9}[((a+b+c)^2-2(ab+bc+ca))^2+(a^2-b^2)^2+(a^2-c^2)^+(b^2-c^2)^2]=\frac{a^4+b^4+c^4}{3}$$
Furthermore we have :
$$\sum_{cyc}^{}(x+y)(y+z)=\sum_{cyc}^{}3xy+y^2$$
And:
$$\sum_{cyc}^{}(x-y)(y-z)=\sum_{cyc}^{}xy-y^2$$
Finally we have :
$$\frac{\sum_{cyc}^{}(x-y)(y-z)+\sum_{cyc}^{}(x+y)(y+z)}{4}=xy+yz+zx$$
And 
$$\frac{1}{3}[(a+b+c)^3-3(a+b)(b+c)(c+a)]=\frac{a^3+b^3+c^3}{3}$$
So the fist identity becomes :
$$\frac{1}{9}[((a+b+c)^2-2(\frac{\sum_{cyc}^{}(a-b)(b-c)+\sum_{cyc}^{}(a+b)(b+c)}{4}))^2+(a^2-b^2)^2+(a^2-c^2)^+(b^2-c^2)^2]=\frac{a^4+b^4+c^4}{3}$$
So if we put the following substitution we get :
$x=a+b$
$y=b+c$
$z=a+c$
We get :
$$\frac{1}{9}[((\frac{x+y+z}{2})^2-2(0.25(xy+zx+zy+(x-y)(y-z)+(y-z)(x-z)+(x-z)(x-y)))^2+x^2(z-y)^2+y^2(x-z)^2+z^2(x-y)^2]=\frac{a^4+b^4+c^4}{3}$$
And 
$$\frac{1}{3}[(\frac{x+y+z}{2})^3-3xyz]=\frac{a^3+b^3+c^3}{3}$$
The condition becomes :
$xyz=8$
Futhermore we now that :
$$x^2(z-y)^2+y^2(x-z)^2+z^2(x-y)^2=2(xy+yz+zx)^2-6xyz(x+y+z)$$
So we have to prove this :
$$\sqrt{[((\frac{x+y+z}{2})^2-2(0.25(2xy+2zx+2zy-(x^2+y^2+z^2)))^2+2(xy+yz+zx)^2-48(x+y+z)]}\leq [(\frac{x+y+z}{2})^3-24] $$
We put $x+y+z=cst=\beta$ and we can maximizing the LHS and the RHS becomes constant . 
We get a one variable inequality this is the following :
$$\sqrt{\frac{11}{48}\beta^4-48\beta}\leq \frac{\beta^3}{8}-24$$
Finally the condition becomes :
$$xyz=8\implies x+y+z\geq 6$$
So the one variable inequality is true for $\beta\geq 6$ 
Done !
