How to prove that $\frac{x^2+y^2}{4}\leq e^{x+y-2}$ I need to prove that given any $x,y\in \mathbb{R}$, such that $x,y\geq 0$, then is true that
$$\frac{x^2+y^2}{4}\leq e^{x+y-2}$$ 
My try was use logarithms and state that without losing generality, if $y\leq x$, I can say that
$$\ln(x^2+y^2)-2\ln(2)\leq \ln(2x^2)-2\ln(2)=2\ln(x)-\ln(2)$$
But I don't see any way to advance more and obtain that this last identity is less or equal to $x+y-2$. Anyone knows how to solve it?
 A: You've already solved it!! You've shown that $\log([x^2+y^2]/4)\leq x+y-2$. You just have to take exponentials on both sides and (because the exponential is an increasing map) you're done.
[It might be noteworthy that your demostration is for the case that $x$ and $y$ are not simultaneously zero, but that case is trivially true]
Edit: Not an answer, yet. I understood @MonsieurGalois had already proved that.
Reedit: It is enough to show that
$$e^{x+y}\geq\frac{x^2+y^2}{4e^2}.$$
As every term in (the power series representation of the exponential map)
$$e^{x+y}=1+(x+y)+\frac{(x+y)^2}{2}+\cdots$$
is positive it follows that $e^{x+y}\geq\frac{(x+y)^2}{2}$, and so that
$$e^{x+y}\geq\frac{(x+y)^2}{2}\geq\frac{x^2+y^2}{2}\geq\frac{x^2+y^2}{4e^2}.$$
Re-reedit: As pointed out by @Clayton, another stupid flaw on me. My apologies.
Because $e^2(x^2+y^2)/4\leq e^2(x+y)^2/4$ it is enough to show that
$$e^2(x+y)^2/4\leq e^{x+y}.$$
This enables us to reduce the question to if $e^2r^2/4\leq e^r$ for $r>0$ (for $r=0$ it is trivially true). Define in that range the function
$$f(r)=4e^r/(er)^2.$$
Our problem is to prove that $f(r)\geq 1$. This value is reached at $r=2$. Furthermore, $f(r)\to+\infty$ when $r\to0$ or $r\to+\infty$. Now we are done cause taking derivatives we get that
$$f'(r)=0\Longleftrightarrow e^rr^2=e^r2r,$$
so $r=2$ is the only critical point of $f(r)$ and given the smothness of $f$ (where it is defined) along with its limits on 0 and $+\infty$ this implies that $r=2$ is the global minimun of $f(r)$.
