<s>[INST] For real numbers $x$ and $y$, show that $\frac{x^2 + y^2}{4} < e^{x+y-2} $ 
Show that for $x$, $y$ real numbers, $0<x$ , $0<y$ 

$$\left(\frac{x^2 + y^2}{4}\right) < e^{x+y-2}.  $$
Someone can help me with this please...
 [/INST] From the well known inequality: $1+z\leq e^z$, we replace $z$ by $\frac{z}{2}-1$ to get $\frac{z}{2}\leq e^{\frac{z}{2}-1}$. Square both sides to get $\frac{z^2}{4}\leq e^{z-2}$. Now let $z=x+y$, to get:  $\frac{(x+y)^2}{4}\leq e^{x+y-2}$. Hence:

$$\frac{x^2}{4}+\frac{y^2}{4}<\frac{(x+y)^2}{4}\leq e^{x+y-2}$$

Note:
$\frac{x^2}{4}+\frac{y^2}{4}< \frac{(x+y)^2}{4}$ because $x,y> 0$
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