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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:

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

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

  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.

  4. 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.

Andreas
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    What is the source of the inequality? Is it related to anything other than other inequalities? – Carl Mummert Mar 19 '18 at 18:47
  • @CarlMummert It doesn' t have a source from some other area (physics, engineering or the like). – Andreas Mar 19 '18 at 18:50
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    I'm not sure if it helps, but note that $$a^4 + b^4 + c^4=(a+b+c)^4 -24(a+b+c)-a^3(b + c)- b^3(a+c) -c^3(a + b)$$ – Surb Mar 19 '18 at 19:34
  • @Andreas I can reduce your inequality to an inequality with one variable based on the identity given by Surb (Thanks to him). Did you want to see that ? –  Mar 21 '18 at 17:07
  • @FatsWallers yes, please. – Andreas Mar 21 '18 at 18:29

4 Answers4

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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.

Nicolas Lussier-Clément
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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!

Michael Rozenberg
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    "Now, let a+b+c be constant and $a^3+b^3+c^3$ be constant." How can you let them be constant? – Andreas Mar 23 '18 at 11:51
  • Why no? But $a^4+b^4+c^4$ changes. – Michael Rozenberg Mar 23 '18 at 14:38
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    You have to give reasons. Homogeneity accounts for one choice, not for two. – Andreas Mar 23 '18 at 14:42
  • Let $a$, $b$ and $c$ be change such that $a+b+c$ and $a^3+b^3+c^3$ they are not change. Thus, $a^4+b^4+c^4$ gets a maximal value for equality case of two variables. – Michael Rozenberg Mar 23 '18 at 14:58
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    "Let a, b and c be change such that a+b+c and $a^3+b^3+c^3$ they are not change." You cannot simply do that, since you remove 2 degrees of freedom. This must be justified. – Andreas Mar 23 '18 at 18:26
  • But we have three variables and $a^4+b^4+c^4$ changes. – Michael Rozenberg Mar 23 '18 at 20:41
  • Again: I argue you are generating a de facto 1-variable problem a priori (!) by introducing two constraints, and this is not justified. – Andreas Mar 24 '18 at 07:10
  • No! Finally $a+b+c$ and $a^3+b^3+c^3$ they are change. But for all given $a+b+c$ and $a^3+b^3+c^3$ we see that $a^4+b^4+c^4$ gets a maximal value, when two variables they are equal. This reasoning not always helps, but in our problem it helps because our inequality depend on $a+b+c$, $a^3+b^3+c^3$ and $a^4+b^4+c^4$ only. – Michael Rozenberg Mar 24 '18 at 07:42
  • It's justified because it solves our problem. – Michael Rozenberg Mar 24 '18 at 07:50
  • I don't agree to the logic: "But for all given a+b+c and $a^3+b^3+c^3$ we see that $a^4+b^4+c^4$ gets a maximal value [...]". Saying "given" means for a priori fixed values, and "For all given" in turn is the same as "for all values", for which the statement is not true. – Andreas Mar 24 '18 at 08:53
  • The statement about the maximal value of $a^4+b^4+c^4$ for equality case of two variables is true. See the Vasc's paper. Your inequality for equality case of two variables is true. Thus, your inequality is proven. It's obvious that we fixed $a+b+c$ and $a^3+b^3+c^3$ for which these values they are exist. – Michael Rozenberg Mar 24 '18 at 09:19
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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 !

  • Dear @FatsWallers I like the approach where you use relations of a Triangle. Can you complete your calculations? – Andreas Apr 04 '18 at 06:39
  • Dear @Andreas I think I can't solve this now but I continue to think about this inequality . Futhermore I add some new details and I think this is profoundly geometric as problem . Thanks again ! –  Apr 05 '18 at 13:51
  • @Andreas I have finally solved your problem . God bless you ! –  Apr 13 '18 at 14:47
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    From your introduction of $\beta = x+y+z$, I identify from the last roots that you have set $((\frac{x+y+z}{2})^2-2(0.25(2xy+2zx+2zy-(x^2+y^2+z^2)))^2+2(xy+yz+zx)^2 = \frac{11}{48}(x+y+z)^4$.

    However, that is not an identity. After expansion, the leading powers of the LHS are $-(x^4 + y^4 + z^4)/8$. So if that is to become any function of $\beta$ it must be $-(x+y+z)^4 /8$. But also that is not true.

    What am I missing?

     – Andreas May 17 '18 at 14:10
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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. $\qquad \Box$

Andreas
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