Inequality involving fractional powers

Question

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?

Answer 1

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 $x,y \geq 0$.