Show: $e^{y^2}-e^{x^2} \le e(y-x)(y+x)$ for $x,y \in [0,1], 0 \lt x \lt y$.

Question

Let $x,y \in [0,1], 0 \lt x \lt y$.

Show: $e^{y^2}-e^{x^2} \le e(y-x)(y+x)$.

I tried to solve it with the mean value theorem:

Let $f: [0,1] \rightarrow \mathbb{R}, \ f(x):=e^{x^2}$, thus $f'(x)=2xe^{x^2}$.

By the mean value theorem, it holds that:

$\exists \lambda \in (0,1): f(y)-f(x) = f'(\lambda)(y-x)$.

It follows:

$$\begin{align} f(y)-f(x) &= e^{y^2}-e^{y^2} \\ &= f'(\lambda)(y-x) = (2\lambda e^{\lambda^2})(y-x) \\ &\le (2e)(y-x) \\ &= e(2y-2x) \\ &= e(\sqrt{2y}-\sqrt{2x})(\sqrt{2y}+\sqrt{2x}). \end{align}$$

I don't know how to go on from there on. I had the idea to split up $e(2y-2x)$ to $e(\sqrt{2y}-\sqrt{2x})(\sqrt{2y}+\sqrt{2x})$ so that I would maybe get to $e(y-x)(y+x)$, but that didn't work out for me.

Any help is appreciated.

Answer 1

Following lessili's comment, I now got:

Let $f:[0,1] \rightarrow \mathbb{R}, \ f(x):=e^x$, thus $f'(x)=e^x$.

By the mean value theorem, it holds that:

$\exists \lambda \in (x^2,y^2): f(y^2)-f(x^2) = f'(\lambda)(y^2-x^2)$.

It follows:

$$f'(\lambda)(y^2-x^2) = e^\lambda (y^2-x^2) \le e(y^2-x^2)=e(y-x)(y+x).$$