$\frac{x^2+y^2}{4}\leq e^{x+y-2}$, in $D=\{(x,y)\in \mathbb R^2| x\in[0,3]\}$

Asked Apr 29 '19 at 18:21

Active Apr 29 '19 at 18:21

Viewed 57 times

I have tried to use convexity properties because I have checked their graphs from (desmos.com) and it was appropriate to try but I am very ignorant to do convexity in two variable so I have tried to look at their differentials.

$$d(e^{x+y-2})=e^{x+y-2}dx+e^{x+y-2}dy$$ $$d\left(\frac{x^2+y^2}{4}\right)=x/2 dx+ y/2 dy$$

I might say that for a fixed $y=y_0$, $e^{x+y-2}dx$ exceed $x/2 dx$ and similarly for $y$ one.

No method comes to my mind, I saw this somewhere and tried to solve but couldnot. Any help, hint would be appreciated.

asked Apr 29 '19 at 18:21

Micheal Brain Hurts

2,380

2

Do you mean $0\leq y \leq 3$ also? For x=0,y=-1 the inequality fails – Amichai Lampert Apr 29 '19 at 19:11
the original was as I wrote but you are right, I guess I should delete the question. – Micheal Brain Hurts Apr 29 '19 at 19:16
Delete the question ? Not necessarily ... You said you have checked with Desmos : what have you checked ? – Jean Marie Apr 29 '19 at 19:34
1

It is correct for $x, y \ge 0$: Show that $\forall (x,y)$ in the first quadrant: $\frac {x^2+y^2}{4}\leq e^{x+y-2}$ – Martin R Apr 29 '19 at 20:16
@Jean Marie. like convexitywise x^2 and x^3 in (0,1) x^2 as exp one is bigger – Micheal Brain Hurts Apr 30 '19 at 02:15

$\frac{x^2+y^2}{4}\leq e^{x+y-2}$, in $D=\{(x,y)\in \mathbb R^2| x\in[0,3]\}$

0 Answers0