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... 
 A: Let $a = 2^x, b = 2^y$. Use some algebraic manipulation to arrive at $(a - 1)^2 + (b - 1)^2 = 2$
This is a circle of radius $\sqrt{2}$ centered at $(1, 1)$, and we want to find the minimum and maximum values of $a + b$.
So $a + b = M \rightarrow b = -a + M$, which is the equation of a line with slope $-1$. The maximum and minimum will be when we find the maximum and minimum y-intercept of this line such that it intersects at least one point on the circle with $a, b > 0$
It's tangent to the circle at the top right when $a = b = 2$ which gives a value of $M = 4$. The smallest y-intercept occurs at $m = 2$, however, we do not include $2$ because it makes one of the coordinates zero. This establishes $(2, 4]$ as the range.
A: So, then the another solution:
let $f(x) = 2^x$. Then we can go to that form of equation:
$$(f(x)-1)^2 + (f(y)-1)^2 = 2$$
We got a circle. If we consider points of intersection of this circle with a lines $$f(x)+f(y) = const$$ we can get an answer.
We should also be polite with restrictions $f(x),f(y) > 0$.
If we draw a picture, we will conclude that range is (2, 4].
A: Cauchy-Schwarz inequality allows us to conclude $$((a-1)^{2}+(b-1)^{2})(1+1) \geq (a-1+b-1)^{2}$$ and hence $4\geq (a+b-2)^{2}$ and finally $a+b \leq 4$ with equality holding only when $a=b=2$.
A: Set $a=2^x$, $b=2^y$, then the problem is equivalent to finding the range of $a+b$ where $a,b>0$ and
$$a^2+b^2=2a+2b$$
Without loss of generality $a\ge b$ and we can make the substitution $c=a+b$, $d=a-b$, so now $c>d\ge0$, and we require
\begin{align*}\left(\frac{c+d}2\right)^2+\left(\frac{c-d}2\right)^2&=(c+d)+(c-d)\\
c^2+2cd+d^2+c^2-2cd+d^2&=8c\\
c^2+d^2&=4c\end{align*}
Observe that
$$c^2\le c^2+d^2=4c\Longleftrightarrow c(c-4)\le0\Longleftrightarrow c\le4$$
and
$$2c^2>c^2+d^2=4c\Longleftrightarrow c(c-2)>0\Longleftrightarrow c>2$$
It's easy to see all such values can be attained and the answer is the interval $(2,4]$.
A: Let us denote $2^{x}=a,2^{y}=b,$where $a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$. 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 $(2,4]$.
A: You have
$$(2^x+2^y)^2=4^x+4^y+2^{x+y+1}=2^{x+1}+2^{y+1}+2^{x+y+1}$$
If $a=2^x$, $b=2^y$, then
$$(a+b)^2=2(a+b+ab)>2(a+b)$$
so $a+b>2$.
I'm looking for an upper bound. If I find it, I'll edit this answer.
A: Let $\delta x = y-x$ and $x\ge y$. Then
$$4^x(1+4^{\delta x}) = 2^{x+1}(1+2^{\delta x})$$
then from factorization we can claim that:
$$2x = x+1 ~\text{and}~~ 4^{\delta x} = 2^{\delta x}$$
So, we get $x = y = 1$.
