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Algebra Difficulty 4.7 AIME Prove it Bulgaria

The reals x,yx, y satisfy x(x−6)≤y(4−y)+7x(x - 6) \le y(4 - y) + 7. Find the minimal and maximal values of the expression x+2yx + 2y.

Solution

Let a=x+2ya = x + 2y. Then x=a−2yx = a - 2y and
(a−2y)(a−2y−6)≤y(4−y)+7a2−2ay−6a−2ay+4y2+12y≤4y−y2+75y2−2(2a−4)y+(a2−6a−7)≤0D=(2a−4)2−5(a2−6a−7)≥04a2−16a+16−5a2+30a+35≥0a2−14a−51≤0(a−17)(a+3)≤0. \begin{aligned} (a - 2y)(a - 2y - 6) &\le y(4 - y) + 7 \\ a^2 - 2ay - 6a - 2ay + 4y^2 + 12y &\le 4y - y^2 + 7 \\ 5y^2 - 2(2a - 4)y + (a^2 - 6a - 7) &\le 0 \\ D &= (2a - 4)^2 - 5(a^2 - 6a - 7) \ge 0 \\ 4a^2 - 16a + 16 - 5a^2 + 30a + 35 &\ge 0 \\ a^2 - 14a - 51 &\le 0 \\ (a - 17)(a + 3) &\le 0. \end{aligned}
Finally, a∈[−3;17]a \in [-3; 17]. □\square

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