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.
 A: Your question is really interesting and this a rather long answer and it is only the reflexion I have so far for this prof. There is still a lot missing to make it complete. Still I decided to post it any way. 
By the way thank for the kindness of all of you on my first attempt. 
Something one can realize is the symmetry of both side of the inequality and on the iso-surface.
$$\frac{a^3+b^3+c^3}{3}$$
$$\sqrt{\frac{a^4+b^4+c^4}{3}}$$ 
$$(a+b)(b+c)(c+a)=8$$ 
When using a geometry approach it easy to see by the plot of the graph that the inequality is true. For that one can plot the iso-surface, $(a+b)(b+c)(c+a)=8$, using polar coordinate, gray in this graph. Then its possible to use the radius to prove the inequality, $a^i=r*a_\phi^i(\theta,\phi)$ where $a^i= (a,b,c)$ 
$$r*a_\phi^i(\theta,\phi)  = (r*a_\phi, r*b_\phi,r*c_\phi)$$ 
By multiply each radius $r(\theta,\phi)$ of the iso-surface by $r_{\neq}$:
$$
r_{\neq}=\frac{a_\phi^3+b_\phi^3+c_\phi^3}
{\sqrt{3(a_\phi^4+b_\phi^4+c_\phi^4)}}
$$
We construct a new iso-surface, red in this graph, and compare there radius. If $r_{\neq}\geq 1$ then the inequality hold true. This is the case. 
To prove it is more difficult as you know. 
The symmetry 
By the fact that $a,b,c$ can be interchange. Symmetry can be describe with 3 plane. Those plane $s_a,s_b,s_c$ are construct with the flowing vector at the origin $(0,0,0)$: $\vec{1}_p=(1,1,1), \vec{1}_a=(0,1,1), \vec{1}_b=(1,0,1), \vec{1}_c=(1,1,0)$
with $s_a=(\vec{1}_p,\vec{1}_a), s_b=(\vec{1}_p,\vec{1}_b), s_a=(\vec{1}_p,\vec{1}_b)$ and the normal to each plane is:
$$
\begin{align*}
\vec{n}_a=(0,-1,1)\\ 
\vec{n}_b=(1,0,-1)\\ 
\vec{n}_c=(-1,1,0)
\end{align*}
$$
Thus the domain can be restricted without lost of generality to one if the 6 symmetric $\mathbb{R}^3$ subset one can get from: 
$$\pm\vec{n}_i\cdot\vec{a} \geq 0$$
where $\vec{a}=(a,b,c)$.
Those 6 symmetric $\mathbb{R}^3$ subset and there boundary of the are describe by any permutation of $a,b,c$ of:
$$
\begin{align}
 a \geq b\geq c\geq 0 \\
 a = b\, \&\, a\geq c \\
 b = c\, \&\, a\geq b \\
 1 \geq c \geq 0
\end{align}
$$
This symmetry old true for the transformation you where using $(a+b)=2x\dots$ and thus on each one of those 6 symmetric region a permutation of $x,y,z$  of these inequality are true by virtue of symmetry. 
$$
\begin{align}
1 \geq y^2z \geq yz^2 \\
x^2z \geq 1 \geq xz^2 \\
x^2y \geq xy^2 \geq 1
\end{align}
$$
I thought that those where interesting to prove the inequality since this is one of the transformation you where talking about. 
On the edge of the symmetric section
There is 6 segment on the surface of (a+b)(b+c)(c+a)=8 that define the boundary of the symmetric region. They can be verified with one equation, on $a=b$ and one on $b=c$ knowing that in this symmetric region $0\leq c \leq1$. One can solve the inequality. 
For the short edge, $a=b$, we have that:
$$ c=\frac{2}{\sqrt{a}}-a$$
And using the same technique to plot the result as for the 3d case we obtain this graph.
For the long edge, $b=c$, we have that:
$$ a=\frac{2}{\sqrt{c}}-c$$
And obtain this graph.  
So far this only prove that there is 6 curve on the edge of the symmetry that respect the inequality. This is a beginning for a geometric understanding of what is going on.  There is a map $g(\theta,\phi)$ where the curve $a=b$, $b=c$ and any other one are one have a fix $\phi$ value where  $\phi \in [0,\pi/3]$ on these curve the inequality is one variable, $\theta$, which is also bound $\theta \in [0,\Theta(\phi)]$ where $\Theta(\phi) \in [0.615,0.955]$. On such a map it might be easier to prove the inequality. But that might be what you are tiring to do. 
A: 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!
A: 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 !
A: 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. $\qquad \Box$
