Find all pairs (x, y) of real numbers such that $16^{x^2+y} + 16^{x+y^2}=1$
Find all pairs (x, y) of real numbers such that
$$16^{x^2+y} + 16^{x+y^2}=1$$
Solution 1:
A very nice problem!
By application of inequality $a+b\ge2\sqrt{ab}$ we have $$16^{x^2+y}+16^{x+y^2}\ge 2\sqrt{16^{x^2+y}\cdot 16^{x+y^2}}=2\cdot 4^{x^2+x+y^2+y}$$ Since $t^2+t\ge-\frac{1}{4}$ holds for all $t\in\mathbb{R}$, for all $x,y\in\mathbb{R}$ we have $$16^{x^2+y}+16^{x+y^2}\ge2\cdot 4^{x^2+x+y^2+y}\ge 2\cdot 4^{-\frac{1}{2}}=1$$ with equality only in case $x=y=-\frac{1}{2}$, so this is the only solution.