Find x + y x+y

Algebra Level 3

1 6 x + y 2 + 1 6 y + x 2 = 1 16^{x+y^2} +16^{y+x^2}=1

If real numbers x x and y y satisfy the equation above, find x + y x+y .


The answer is -1.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Mark Hennings
Feb 25, 2020

Note that 1 2 = 1 6 x + y 2 + 1 6 y + x 2 2 ≥ 1 6 x + y 2 + y + x 2 = 4 ( x + 1 2 ) 2 + ( y + 1 2 ) 2 − 1 2 = 1 2 × 4 ( x + 1 2 ) 2 + ( y + 1 2 ) 2 \frac12 = \frac{16^{x+y^2} + 16^{y+x^2}}{2} \;\ge \; \sqrt{16^{x+y^2 + y+x^2}} \; = \; 4^{(x+\frac12)^2 + (y+\frac12)^2 - \frac12} \; = \; \frac12 \times 4^{(x+\frac12)^2 + (y+\frac12)^2} using the AM/GM inequality, so that 4 ( x + 1 2 ) 2 + ( y + 1 2 ) 2 ≤ 1 4^{(x+\frac12)^2 + (y+\frac12)^2} \le 1 which implies that x = y = − 1 2 x=y=-\tfrac12 , so that x + y = − 1 x+y=\boxed{-1} .

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