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Question Number 204920 by cortano12 last updated on 02/Mar/24

$${16}^{\mathrm{y}+{\mathrm{x}}^2}+{16}^{{\mathrm{y}}^2+\mathrm{x}}=1$$$$\mathrm{x}+\mathrm{y}=?$$

Answered by witcher3 last updated on 02/Mar/24

$${16}^{\mathrm{y}+{\mathrm{x}}^2}+{16}^{{\mathrm{y}}^2+\mathrm{x}}=1\Rightarrow \mathrm{AM}-\mathrm{GM}$$$$2\sqrt{{16}^{\mathrm{y}+{\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}}}⩽1$$$$\iff 2.2^{2(\mathrm{y}+{\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x})}⩽2^0$$$$\mathrm{y}+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{y}}^2⩽-\frac{1}{2}$$$${(\mathrm{x}+\frac{1}{2})}^2+{(\mathrm{y}+\frac{1}{2})}^2⩽0$$$$\Rightarrow (\mathrm{x}=-\frac{1}{2};\mathrm{y}=-\frac{1}{2})$$$${\mathrm{x}}^2+\mathrm{y}={\mathrm{y}}^2+\mathrm{x}=-\frac{1}{4}\iff {16}^{-\frac{1}{4}}+{16}^{-\frac{1}{4}}=1\mathrm{True}$$$$\mathrm{x}+\mathrm{y}=-1\mathrm{\&}(\mathrm{x}=\mathrm{y})$$$$$$$$$$$$$$$$$$

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