<s>[INST] Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$

I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question... 
 [/INST] Let us denote $2^{x}=a,2^{y}=b,$where $a,b\in 
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%BeginExpansion
\mathbb{R}
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^{+}$. Then we have $a^{2}+b^{2}=2(a+b)$ which is equivalent to $%
(a-1)^{2}+(b-1)^{2}=2.$So we take the part of the circle  with center $M(1,1)
$ and radius $r=\sqrt{2}$ in the first district of the plane as the domain of the
function $f(a,b)=a+b$. We are looking for maxium and minium values of $f$ in
the given domain. Any point $(a,b)$ in the domain is $\sqrt{a^{2}+b^{2}}$
units far from $(0,0)$ which is also in the given circle (though not in
domain). We also know that two points on a circle are at most diameter
length far from each other. So $\sqrt{a^{2}+b^{2}}=2\sqrt{2}$ at most. Thus $%
a^{2}+b^{2}=8.$ Then we have $2(a+b)=8$ and finally $a+b=4$ at maximum. We
look at the edges (because they are closer to (0,0) ) of the domain for
mininum value which is at $(2,0)$ or $(0,2)$. Then $a+b$ converges to $2$ at
minimum. We have the value range $(2,4]$.
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