For real numbers $x$ and $y$, show that $\frac{x^2 + y^2}{4} < e^{x+y-2} $

Question

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...

Answer 1

From the well known inequality: $1+z\leq e^z$, we replace $z$ by $\frac{z}{2}-1$ to get $\frac{z}{2}\leq e^{\frac{z}{2}-1}$. Square both sides to get $\frac{z^2}{4}\leq e^{z-2}$. Now let $z=x+y$, to get: $\frac{(x+y)^2}{4}\leq e^{x+y-2}$. Hence:

$$\frac{x^2}{4}+\frac{y^2}{4}<\frac{(x+y)^2}{4}\leq e^{x+y-2}$$

Note:

$\frac{x^2}{4}+\frac{y^2}{4}< \frac{(x+y)^2}{4}$ because $x,y> 0$

Answer 2

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.

Answer 3

This is only a partial answer because it works only for $x+y>2$, but it might give you some insight:

First , we need to reduce this to one variable, so let's use the inequality $\sqrt{a+b}<\sqrt{a}+\sqrt{b}$ (this one is easily proved). We take the square root of both sides, and then apply this inequality to the LHS:

$\frac{x+y}{2}<\sqrt{e^{x+y-2}}$

Let $r=x+y$ , and let's rewrite this:

$\frac r2 <{e^\frac r2 -1}$ (I don't know why it's not showing it but the $-1$ is in the exponent from now on)

If we take the derivative of both sides with respect to $r$ variables, we see that we need to prove that

$\frac 12 < \frac 12 {e^\frac r2 -1}$ , or $1<{e^\frac r2 -1}$

I'm sure you already know that exponential functions are monotonically increasing, so we don't need to go over that. If you solve ${e^\frac r2 -1}>1$ , you get $r>2$