# Is the following inequality true: $x^{t}y^{1-t} \geq tx+(1-t)y$ for $t\in [0,1], x,y > 0$?

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.

• 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.
– user593746
Oct 9, 2018 at 14:29

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.
• Perhaps you meant $\sqrt{xy}\ge\frac{x+y}{\color{#C00}{2}}$.
$$x=0.1,y=1,t=0.5$$ is a counterexample. Left hand side is about $$0.31$$, the right hand side is $$0.55$$
$$4^{1/2}9^{1-1/2}\ge \frac12\cdot 4+\left(1-\frac12\right)9?$$