Suppose that we have two vectors, $x,y\in\mathbb{R}^n$. Let $z\in\mathbb{R}^n$ be the vector given by $z_i=\sqrt{x_i y_i}$. With an abuse of notation, I may write $z=\sqrt{xy}$. Consider the quantity $X(t)=2\|z\|_{1}\|z\|_{t}^{t}-\left(\|x\|_{1}\|y\|_{t}^{t}+\|y\|_{1}\|x\|_{t}^{t}\right)$ where $\|x\|_t=\left(x_1^t +\cdots+x_n^t\right)^\frac{1}{t}$ refers to the $t$-norm.
I only care about when $t$ is very very close to $t=1$. Notice that $X(1)=2\|z\|_1^2-\|x\|_1\|y\|_1=2\left(\sum_{i=1}^n \sqrt{x_i y_i}\right)^2-2\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right).$ We can then show that $X(1)\leq 0$ by expanding and using the AM_GM.
Here is my question:
Is it possible to show that for very small $\epsilon$ we have, $X(1-\epsilon)\leq X(1)\leq X(1+\epsilon)?$