<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] When $x+y$ is fixed $x^2+y^2=(x+y)^2-2xy$ takes maximal value when $xy=0.$ So it is enough to prove our inequality when one variable is $0.$ In this case, it can be reduced to $\frac{x^2}{4}\le e^{x-2}$ or $e^z\ge 1+z+\frac{z^2}{4}$ which immediately follows from the fact that $e^z\ge 1+z+\frac{z^2}{2}.$
Alternative way to present solution:  Let $x+y=t$
$$\frac{x^2+y^2}{4}=\frac{(x+y)^2-2xy}{4}\le \frac{t^2}{4}\le e^{t-2}$$
and the rest is the same as above.
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