Question Number 151806 by liberty last updated on 23/Aug/21
Commented by ghimisi last updated on 23/Aug/21
$${x}={y}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by ghimisi last updated on 23/Aug/21
$$\mathrm{36}^{{x}^{\mathrm{2}} +{y}} +\mathrm{36}^{{y}^{\mathrm{2}} +{x}} \overset{{am}−{gm}} {\geqslant}\mathrm{2}\sqrt{\mathrm{36}^{{x}^{\mathrm{2}} +{y}+{y}^{\mathrm{2}} +{x}} } \\ $$$$=\mathrm{2}\centerdot\mathrm{6}^{{x}^{\mathrm{2}} +{x}+{y}^{\mathrm{2}} +{y}} \geqslant\mathrm{2}\centerdot\mathrm{6}^{−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}\Rightarrow \\ $$$${x}^{\mathrm{2}} +{y}={x}^{\mathrm{2}} +{y} \\ $$$${x}^{\mathrm{2}} +{x}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}^{\mathrm{2}} +{y}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}={y}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$