Show that $\forall (x,y)$ in the first quadrant: $\frac {x^2+y^2}{4}\leq e^{x+y-2}$ I have the folowing exercise (which I've been thinking quite a while and couldn't figure out):
Show that $\forall (x,y)$ in the first quadrant:
$$\frac {x^2+y^2}{4}\leq e^{x+y-2}$$
My idea was to work with maxima and minima, but I'm stuck...
Any help will be much appreciated!
 A: Look at this on a given circle $x^2 + y^2 = r^2$.. for which $\theta$ is the right-hand side smallest? This will reduce it to a problem involving only one variable.
A: Taking @Zarrax's suggestion, you will find that the minimum value of the exponent is $r-2$, where $r=\sqrt{x^2+y^2}$.  Taking logs of both sides, you get $2 \log{(r/2)}$ for the LHS and $r-2$ for the RHS.  Now apply the inequality $\log{x} \le x-1 \: \forall \, x>0$ where $x=r/2$, and the inequality follows.
A: (Essentially the same idea as in the previous  answers, just put slightly differently:)
For $x, y \ge 0$
$$
 \frac {x^2+y^2}{4} \le \frac {(x+y)^2}{4} = \left( \frac{x+y}{2} \right)^2 \, ,
$$
therefore it suffices to show that
$$
\left( \frac{x+y}{2} \right)^2 \le e^{x+y-2} \Longleftrightarrow 
\frac{x+y}{2} \le e^{\frac{x+y}{2}-1} \, .
$$
The latter is true because 
$$
 1 + u \le e^{u} 
$$
holds for all real numbers $u$ (the right-hand side is a convex function and
therefore its graph lies above the tangent line at $u=0$, 
see also Simplest or nicest proof that $1+x \le e^x$).
