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Question-151806




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

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