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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!

ivan
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HipsterMathematician
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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 1591719 Jan 25 '13 at 15:44

3 Answers3

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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.

Zarrax
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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
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(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 $1+x \le e^x$).

Community
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Martin R
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  • I think it comes out to $eu \le e^u$. Were you implying that it's very easy to prove this from $1 + u \le e^{u}$? – Ovi Nov 25 '18 at 05:09
  • @Ovi: With the substitution $u= \frac{x+y}{2}-1$ the last inequality becomes $u+1 \le e^u$. – Martin R Nov 25 '18 at 15:22
  • Ahhhh thank you – Ovi Nov 25 '18 at 16:40
  • That is so cool/weird that you can go back and forth between an additive inequality, $1+x \le e^x$ and $ex \le e^{x-1}$. I hope I will remember this trick and be able to use it in other cases. Is this an important trick that you keep in mind, or come from a more general method, or is it just a cute little thing? – Ovi Nov 25 '18 at 17:23
  • @Ovi: Going from additive (in)equalities to multiplicative (in)equalities and vice versa, via $e^{x+y} = e^x e^y$ or $\log (xy) = \log(x) + \log(y)$, is surely something that you should keep in mind. – Martin R Nov 25 '18 at 17:42
  • To go from $1+x \le e^x$ to $ey \le e^{y-1}$ we use $x =y-1$. Could you please clarify how you would use $e^{x+y} = e^x e^y$ to switch between additive and multiplicative? I of course see addition on one side and multiplication on the other, but I don't see how it would be used in the same way that we used $x =y-1$. – Ovi Nov 25 '18 at 17:48
  • @Ovi: It uses $e^{y-1} = e^y / e^1$ ... – Martin R Nov 25 '18 at 18:23
  • Yes, I just thought you meant there was something more to it than that. Thanks for all the help! – Ovi Nov 25 '18 at 18:35
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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.

Ron Gordon
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