Puzzled by this still open question, I tried comparing the arithmetic mean $A(x,y)=(x+y)/2$ with a mean intermediate between a geometric-type mean $G(X)=(x^a y^{1-a}+x^{1-a} y^a)/2\;$ for $0\le a \le 1$, and an $L^p$-type mean $L(x,y)=(x^p+y^p)^{1/p}\;$ for $p\ge 1$.

Notice that $G(x,y)\le A(x,y)\le L(x,y)\;$ for $x\ge0, y\ge0$. Then define the mixed mean

$$M_{a,p}(x,y)=\Big(\frac{(x^a y^{1-a})^p+(x^{1-a} y^a)^p}{2}\Big)^{\frac{1}{p}}$$

The following are easily proved:

  • $M_{a,p}\le M_{b,q}\;$ if $\;1\le p\le q\;$ and $\;1/2\le a\le b\;$;

  • given $1/2\le a< 1$ and $p\ge1$, then $\;A(x,y)>M_{a,p}(x,y)\;$ for $x/y$ or $y/x$ large enough;

  • the Hessian of $\;M_{a,p}\;$ at $\;(x,y)=(1,1)\;$ is $\;\displaystyle\frac{p(p(2a-1)^2-1)}{8}\left( \begin{smallmatrix} 1&-1\\ -1&1\end{smallmatrix}\right)\;$, which is negative semidefinite for $p\le (2a-1)^{-2}\;$;


$$M_{a,p}\le A=M_{1,1}\;\; \text{on}\;\; \mathbb{R}_{+}^2\implies p\le \frac{1}{(2a-1)^2}$$

Question: is the converse true? In other words, is $M_{a,\frac{1}{(2a-1)^2}}\le A\;$?

After checking few numerical examples, I would guess a positive answer.

Notice that the more general comparison between $M_{a,p}$ and $M_{b,q}$ can be reduced easily to the comparison between some other $M_{c,r}$ and $M_{1,1}$.

In 4 variables, the local condition implied by the Hessian is not sufficient (see example here).


Let me change the notation a bit by calling $m=\frac{1}{2a-1}>1$. Then your inequality reads $$\left(\frac{x+y}{2}\right)^{m^2}\geq (xy)^{\binom{m}{2}}\cdot\left(\frac{x^m+y^m}{2}\right).$$ This is true for all $x,y\geq 0$, and I would love to see a slick elementary proof. However in the meantime notice that you can prove it by using Lagrange multipliers. Let's normalize to $x+y=1$ and check that the critical points of $(xy)^{\binom{m}{2}}\left(x^m+y^m\right)$ happen where $$(m-1)(x^{m+1}-y^{m+1})=(m+1)xy(x^{m-1}-y^{m-1}),$$ but it's not hard to check that this can only happen if $x=y$. And so by a simple check on the boundary values, one can conclude that your inequality is true.

  • $\begingroup$ It may not be so elementary... but it certainly is slick! $\endgroup$ – Yaakov Baruch Jan 22 '15 at 12:52
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    $\begingroup$ If you normalise instead to $xy=1$ and write $x = e^t$, the inequality becomes $\cosh(t)^{m^2} \geq \cosh(mt)$. It suffices then to show that $(\log \cosh(\sqrt{x})) / x$ is decreasing on the positive real axis, which I can confirm numerically but don't have a conceptual proof. $\endgroup$ – Terry Tao Jan 22 '15 at 22:01
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    $\begingroup$ @TerryTao: Curiously, checking $\log(\cosh(x))/x^2$ is decreasing also showed up here, though again not without calculus: mathoverflow.net/questions/40334/… $\endgroup$ – Suvrit Jan 28 '15 at 16:37

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