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Question Number 151806 by liberty last updated on 23/Aug/21

Commented by ghimisi last updated on 23/Aug/21

$$x=y=\frac{-1}{2}$$$$$$

Answered by ghimisi last updated on 23/Aug/21

$${36}^{x^2+y}+{36}^{y^2+x}\stackrel{am-gm}{⩾}2\sqrt{{36}^{x^2+y+y^2+x}}$$$$=2\cdot 6^{x^2+x+y^2+y}⩾2\cdot 6^{-\frac{1}{4}-\frac{1}{4}}=\frac{2}{\sqrt{6}}\Rightarrow$$$$x^2+y=x^2+y$$$$x^2+x=-\frac{1}{4}$$$$y^2+y=-\frac{1}{4}$$$$x=y=-\frac{1}{2}$$

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