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Algebra Difficulty 4.7 AIME Find the answer

5. Given x,yR+x, y \in \mathbf{R}_{+}. Then the maximum value of x2x+y+yx+2y\frac{x}{2 x+y}+\frac{y}{x+2 y} is ( ).

A number or a short expression. Spacing and $ signs are ignored.

Solution

5. B.

Let s=2x+y,t=x+2ys=2x+y, t=x+2y. Then
x=13(2st),y=13(2ts). \begin{array}{l} x=\frac{1}{3}(2s-t), \\ y=\frac{1}{3}(2t-s) . \end{array}
Thus x2x+y+yx+2y=4313(ts+st)23 \text{Thus } \frac{x}{2x+y}+\frac{y}{x+2y}=\frac{4}{3}-\frac{1}{3}\left(\frac{t}{s}+\frac{s}{t}\right) \leqslant \frac{2}{3} \text{. }

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.