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

Question

Show that \begin{equation} \frac{x^2 + y^2}{4} \leq e^{x+y-2} \end{equation} is true for $x,y \geq 0$.

As far, I have prove that \begin{equation} x^2 + y^2 \leq e^{x}e^{y}\leq e^{x+y} \end{equation} since $e^{x}\geq x^2$ and $e^{y}\geq y^2$. If someone can give some aid it would be nice!

Answer 1

Use that $e^t\geq 1+t$ for all $t\in\mathbb{R}$ and $x+y\geq 0$: $$e^{x+y-2}=\left(e^{\frac{x+y}{2}-1}\right)^2\geq\left(\frac{x+y}{2}\right)^2.$$

Answer 2

To sketch it out :

Let $x = r\cos\theta $ and $y=r\sin\theta$. Upon considering both $x$, $y$ are $\geq 0$ we get $\theta \in [0,\frac{\pi}{2}]$ and $r>0$. Plug in the substitutions into your initial inequality to get $\frac{e^2r^2}{4} \leq e^{r(\cos\theta+\sin\theta)}$ .

Now, on $[0,\frac{\pi}{2}]$ we get (pretty intuitively) that $\min(\sin\theta +\cos\theta) = 1$. Use this fact, along with a truncated version of the Taylor expansion for $e^x$ to complete your proof.