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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!

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Does it help to denote $$ f(x,y)=e^{x+y-2}-\frac {x^2+y^2}{4}$$ and then find the minima of the function in the first quadrant by following the multivariable tools? It seems to me the natural way to follow. – user 1618033 Jan 25 '13 at 15:44

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.

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Could you explain a bit better? I'm almost there.... – HipsterMathematician Jan 25 '13 at 14:53
write $x = r\cos(\theta)$ and $y = r\sin(\theta)$. The left hand side is ${r^2 \over 4}$, and the right hand side is a function of $r$ and $\theta$. Minimize the right-hand side in $\theta$, and now you have a 1-dimensional problem (which still requires some work) – Zarrax Jan 25 '13 at 14:58

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.

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