This is a question in vietnamese national math exam at the end of 12th grade.
Given x,y are real numbers which satisfy the condition: $x+y+1=2(\sqrt{x-2}+\sqrt{y+3})$
Find a $m$ such that:
$3^{x+y-4}+(x+y+1)2^{7-x-y}-3(x^2+y^2)\leq m$
The answer as given is $m \geq 148/3$, which contradicts with wolfram alpha (it claims that it has no global maximum). I'm unable to find a counter example due to the non-linear nature of the system of equations.
I hope that math SE could help.
Note: This is NOT an answer. This is just a long comment with some thought.
Let $u=\sqrt{x-2}, v=\sqrt{y+3}\implies x=u^2+2,y=v^2-3$
We have $$u^2+2+v^2-3+1=2(u+v)$$ $$(u-1)^2+(v-1)^2=2$$
Set $u=\sqrt2 \sin t+1,v=\sqrt2 \cos t+1$, $t\in[\frac{\pi}4,\frac{3\pi}4]$
Substitute back to original equation:
\begin{align} x+y+1&=(\sqrt2 \sin t+1)^2+2+(\sqrt2 \cos t+1)^2-3+1\\&=4+2\sqrt2(\sin t+\cos t) \end{align}
Denote $L=x+y+1$.
Extremum of $L$ is obtained when $t=\frac{\pi}4$. (By differentitation) We get $L=8.$
Back to the inequality,
$$3^{L-5}+L\cdot2^{8-L}-3(x^2+y^2)\le m$$