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I encountered this problem when trying to determine whether the objective function of some problem is convex. I was able to reduce the proof to the above inequality.

I have not been able to find a counter example to the inequality so I suspect it is probably true. However, I am completely stuck on how I would go about proving that this is indeed the case. Any help would be very appreciated.

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    $\begingroup$ The reversed inequality is true. It is a generalized version of AM-GM for two variables. That is, $$x^ty^{1-t}\leq tx+(1-t)y$$ for all $t\in[0,1]$ and $x,y>0$. The equality cases are $t=0$, $t=1$, and $x=y$. You can prove this by Jensen's inequality or just differentiation. $\endgroup$
    – user593746
    Oct 9, 2018 at 14:29

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The converse is true: $$x^{t}y^{1-t} \leq tx+(1-t)y$$ since, as the logarithm is a concave function, $$\log(x^{t}y^{1-t})=t\log x+(1-t)\log y \leq \log\bigl(tx+(1-t)y\bigr).$$ Note: If your inequality were true, it would imply $\sqrt{xy}\ge \dfrac{x+y}2$, which contradicts the AGM inequality.

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  • $\begingroup$ Perhaps you meant $\sqrt{xy}\ge\frac{x+y}{\color{#C00}{2}}$. $\endgroup$
    – robjohn ♦
    Oct 9, 2018 at 15:47
  • $\begingroup$ @robjohn: Oops. I should reread what I did write before posting! Thanks for pointing the typo! $\endgroup$
    – Bernard
    Oct 9, 2018 at 16:10
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$x=0.1,y=1,t=0.5$ is a counterexample. Left hand side is about $0.31$, the right hand side is $0.55$

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Is

$$4^{1/2}9^{1-1/2}\ge \frac12\cdot 4+\left(1-\frac12\right)9?$$

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