Prove estimation of inequalities

Question

How can I show the inequality $e^{y^2}-e^{x^2}\leq e(y-x)(y+x) \text{ for } x,y\in [0,1] \text{ with } 0<x<y$

and

$(1+x-y)\cdot e^{\frac{1}{x}}\leq e^{\frac{1}{y}} \text{ for } x,y\in\mathbb{R} \text{ with } 1<x<y$?

Thanks in advance!

Answer 1

For the first, the mean value theorem for $e^t$ on the interval $[x^2,y^2]$ gives you $$\frac{e^{y^2}-e^{x^2}}{y^2-x^2}=e^t $$ for some $t\in(x^2,y^2)$. What can you say about $e^t$ given that $x^2<t<y^2\leq 1$?

For the second, rewrite it as $$e^{\frac 1y-\frac 1x}\geq 1+x-y $$ Then use the well-known inequality $e^t\geq t+1$ for all $t\in\mathbb{R}$. $$e^{\frac 1y-\frac 1x}\geq \frac 1y-\frac 1x+1 $$ So you have to show $$\frac 1y-\frac 1x+1 \geq 1+x-y \Leftrightarrow \\ \frac{x-y}{xy}\geq x-y $$ for $1<x<y$.

Answer 2

Hint

The inequality is $$1+x-y\le e^{{1\over y}-{1\over x}}=e^{x-y\over xy}$$Define $u=x-y<0$ and show that $$1+u\le e^{\alpha u}\quad,\quad \forall u\le 0,\forall 0<\alpha<1$$and substitute $\alpha ={1\over xy}$.

Answer 3

A variant for the second inequality:

Rewrite if as $$\mathrm e^{\tfrac1x}-\mathrm e^{\tfrac1y}\le (y-x) \mathrm e^{\tfrac1x}\qquad\text{if }\;1<x<y.$$ Apply the Mean Value theorem to the exponential function: there exists a real number $c$ such that $$\mathrm e^{\tfrac1x}-\mathrm e^{\tfrac1y}=\Bigl(\frac 1x -\frac 1y\Bigr)\mathrm e^c=\frac{y-x}{xy}\,\mathrm e^c,\qquad \frac1y<c<\frac 1x. $$ Now observe that $\;c<\frac 1x < 1\;$ and $\;xy>1$, so $$\mathrm e^{\tfrac1x}-\mathrm e^{\tfrac1y}=\Bigl(\frac 1x -\frac 1y\Bigr)\mathrm e^c=\frac{y-x}{xy}\,\mathrm e^c<(y-x)\mathrm e^{\tfrac 1x}.$$