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.

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    $\begingroup$ Do you mean $0\leq y \leq 3$ also? For x=0,y=-1 the inequality fails $\endgroup$ – Amichai Lampert Apr 29 at 19:11
  • $\begingroup$ the original was as I wrote but you are right, I guess I should delete the question. $\endgroup$ – user2312512851 Apr 29 at 19:16
  • $\begingroup$ Delete the question ? Not necessarily ... You said you have checked with Desmos : what have you checked ? $\endgroup$ – Jean Marie Apr 29 at 19:34
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    $\begingroup$ It is correct for $x, y \ge 0$: Show that $\forall (x,y)$ in the first quadrant: $\frac {x^2+y^2}{4}\leq e^{x+y-2}$ $\endgroup$ – Martin R Apr 29 at 20:16
  • $\begingroup$ @Jean Marie. like convexitywise x^2 and x^3 in (0,1) x^2 as exp one is bigger $\endgroup$ – user2312512851 Apr 30 at 2:15

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