# Inequality involving fractional powers

I am interested in the lower bound of the following quantity $$2^{1+\epsilon}(x^{1+ 2\epsilon }+y^{1+ 2\epsilon})-(x^{\epsilon}+y^{\epsilon})(x+y)^{1+ \epsilon}$$, where $$x,y, \epsilon \in [0,1]$$ and $$x+y \leq 1$$.

My hunch is the above quantity is always non-negative. For example, I can prove that for $$\epsilon=1$$ it is non-negative for all $$x, y$$. For $$\epsilon=0$$ it equals $$0$$. For $$|x-y| =1$$ it is positive and for $$x=y$$ it is $$0$$.

Can anyone please provide a lower bound (non-negative ??) proof?

Consider the function $$f : [0, 1] \to \mathbb{R}$$: $$f(t) = (1-t)^k + t^k$$ One can check that $$f$$ is convex for $$k \geq 1$$, and concave for $$0 < k < 1$$, with $$t = \frac{1}{2}$$ as the local minimum/maximum. This implies that for $$k \geq 1$$, $$f(t) \geq \left(\frac{1}{2}\right)^k + \left(\frac{1}{2}\right)^k = 2\left(\frac{1}{2}\right)^k$$, and for $$0 < k < 1$$ we have $$f(t) \leq 2\left(\frac{1}{2}\right)^k$$. I'll leave you to check that this implies: $$x^{1+2\epsilon} + y^{1+2\epsilon} \geq 2\left(\frac{x+y}{2}\right)^{1+2\epsilon} \\ x^{\epsilon} + y^{\epsilon} \leq 2\left(\frac{x+y}{2}\right)^{\epsilon}$$ Therefore: $$2^{1+\epsilon}(x^{1+ 2\epsilon }+y^{1+ 2\epsilon})-(x^{\epsilon}+y^{\epsilon})(x+y)^{1+ \epsilon} \geq 2^{2+\epsilon}\left(\frac{x+y}{2}\right)^{1+2\epsilon} - 2\left(\frac{x+y}{2}\right)^{\epsilon}(x+y)^{1+\epsilon} = 0$$ This inequality applies for all $$x,y \geq 0$$.