Let $a, b, c$ be positive real numbers, show that

$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$

I think this is likely to turn out to be proved by Hölder, but I can't see how. Any hints will be appreciated.

  • $\begingroup$ One way is to do this in steps, $\text{LHS}\ge\cdots\ge\cdots\ge\text{RHS}$, where each step is simpler. You can bound RHS from above more easily if you assume $a+b+c=1$, while the LHS could be bounded from below if you assume $a^2+b^2+c^2=1$. Note that the RHS is "concave-like" in the sense that it is maximal "in the middle" where $a=b=c$, while the LHS is maximal along the edges: this statements makes sense as both sides are homogeneous (of degree 4). So you have quite a bit of leeway. $\endgroup$ – Einar Rødland Apr 17 '16 at 8:30
  • $\begingroup$ @EinarRødland I don't quite understand you? Can you explain in detail? $\endgroup$ – Colescu Apr 17 '16 at 8:32
  • $\begingroup$ That would almost require giving you the solution. I'll give a short version in an answer. $\endgroup$ – Einar Rødland Apr 17 '16 at 8:34

W.L.O.G., let $a\ge b\ge c$, then by the rearrangement inequality, we have \begin{align*} \frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2} &\ge \frac{b^6}{b^2+c^2}+\frac{c^6}{a^2+c^2}+\frac{a^6}{a^2+b^2},\tag{1}\\ \frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2} &\ge \frac{c^6}{b^2+c^2}+\frac{a^6}{a^2+c^2}+\frac{b^6}{a^2+b^2}.\tag{2} \end{align*} Thus combine $(1)$ and $(2)$, \begin{align*} 2&\left(\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}\right)\\ &=\frac{b^6+c^6}{b^2+c^2}+\frac{a^6+c^6}{a^2+c^2}+\frac{a^6+b^6}{a^2+b^2}\\ &=(b^4-b^2c^2+c^4)+(a^4-a^2c^2+c^4)+(a^4-a^2b^2+b^4)\\ &=(a^4+b^4+c^4)+(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2)\\ &\ge a^4+b^4+c^4\tag{3}\\ &=\frac{a^4+b^4}{2}+\frac{b^4+c^4}{2}+\frac{c^4+a^4}{2}\\ &\ge \frac{a^4+b^2c^2}{2}+\frac{b^4+a^2c^2}{2}+\frac{c^4+a^2b^2}{2}\tag{4}\\ &\ge a^2bc+b^2ac+c^2ab\\ &=abc(a+b+c), \end{align*} where $(3)$ and $(4)$ are also used the rearrangement inequality. So the result follows.

  • $\begingroup$ Brilliant! Still, can it be proved from Hölder's inequality? $\endgroup$ – Colescu Apr 17 '16 at 8:49
  • $\begingroup$ I did'nt think that yet, I'll try it later. $\endgroup$ – Solumilkyu Apr 17 '16 at 8:52
  • $\begingroup$ The inequality is cyclic, but not symmetric in $a, b, c$, so why can you assume that $a \ge b \ge c$? $\endgroup$ – Martin R Apr 17 '16 at 9:00
  • $\begingroup$ @MartinR Does it have to be symmetric? Why? $\endgroup$ – Colescu Apr 17 '16 at 9:04
  • $\begingroup$ @MartinR I think the fact that inequalities $(1)$ to $(4)$ hold is independent to the choice of the order between $a$, $b$, and $c$. $\endgroup$ – Solumilkyu Apr 17 '16 at 9:06

Note that both sides are homogeneous of degree 4: ie, if we replace $a,b,c$ by $ta,tb,tc$ where $t>0$, both sides get multiplied by $t^4$ which has no effect on the inequality.

If we rescale by $t=1/(a+b+c)$, the effect is the same as requiring that $a+b+c=1$. Often this is formulated as the statement that we can choose $a+b+c=1$ without loss of generality. Alternatively, we could have chosen $a^2+b^2+c^2=1$ without loss of generality, which would be convenient for reformulating the LHS. Unfortunately, we cannot get both these assumptions at the same time. However, we can use then in a step-wise manner to break the inequality into three steps:

$$ \sum_{\text{cycle}(a,b,c)}\frac{a^6}{b^2+c^2} \ge\frac{(a^2+b^2+c^2)^2}{6} \ge\frac{(a+b+c)^4}{2\cdot3^3} \ge\frac{abc(a+b+c)}{2} $$

The way I got to this was the realisation that the RHS expression could be bounded from above more easily under the assumption $a+b+c=1$. Basically, if $a+b+c=1$, we have $abc\le1/3^3$, and so $RHS\le1/(2\cdot3^3)$. However, this inequality now only applies to $a+b+c=1$, so the return to the general case of any $a,b,c>0$, multiply by $(a+b+c)^4$, which is just $1$ when $a+b+c=1$ to make it $$ \frac{abc(a+b+c)}{2}\le\frac{1}{2\cdot3^3}=\frac{(a+b+c)^4}{2\cdot3^3} \quad\text{when}\quad a+b+c=1. $$ However, the outermost terms, $abc(a+b+c)/2\le(a+b+c)^4/(2\cdot3^3)$ is now homogeneous of degree 4 in $a,b,c$ and so holds for all $a,b,c>0$.

The LHS can be made easier to address if we assume $a^2+b^2+c^2=1$. Then, $a^6/(b^2+c^2)=a^6/(1-a^2)$. On this we can apply Jensen's inequality, but in terms of $u=a^2$, $v=b^2$, $w=c^2$ with $u+v+w=1$. Again, we work under the restriction $u+v+w=1$, and then multiply in the appropriate power of $u+v+w$ at the end to get homogeneous expressions of the same degree.

After the two simplifying bounds, we get the inequality in the middle of the topmost chain of inequalities, which shouldn't be too hard.

  • $\begingroup$ Do you mean $\sum_{\text{cycle}(a,b,c)} \frac{a^6} {b^2+c^2} \ge \frac {(a^2+b^2+c^2)^2}{6} \ge \frac {(a+b+c)^4}{2\cdot3^3} \ge \frac {abc(a+b+c)}{2}$? Was it a typo? $\endgroup$ – Colescu Apr 17 '16 at 9:02
  • $\begingroup$ Yes, that was a typo. Fixed now. $\endgroup$ – Einar Rødland Apr 17 '16 at 9:05
  • $\begingroup$ This is way simpler and more elegant than the one by rearrangement. Thank you! $\endgroup$ – Colescu Apr 17 '16 at 9:06

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