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Knowing that $9a^2+8ab+7b^2 \le 6$, prove that $7a+5b+12ab \le 9$. I have found the same question here, but the answer looks wrong. You can't just add two inequalities like him. It's like you would multiply the second one with $-1$, which changes its sign, and then add them.

Here is my try: $$9a^2 + 9b^2 + 18ab - 2b^2 - 10ab - 6 \le 0 \Longleftrightarrow\\ (3a+3b)^2 - 2(b^2+5ab+3) \le 0 \Longleftrightarrow\\ b^2+5ab+3 \ge 0 \Longleftrightarrow\\ 25a^2 - 12 \le 0 \Longleftrightarrow\\ a \in \left[-\frac{2\sqrt{3}}{5}, \frac{2\sqrt{3}}{5} \right]$$

gareth618
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  • But this would effectively be a duplicate. If you want a better answer, please put a bounty on the old question instead... Oh wait, unfortunately that question is closed. – user061703 Jul 17 '18 at 12:08
  • Exactly, it's closed. – gareth618 Jul 17 '18 at 12:10
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    The point in the closed question is that a sum of squares is always non-negative. The answer picks a sum of squares which is relevant to the question and leads to a solution. The answer also tries to explain how that particular sum of squares was discovered. – Mark Bennet Jul 17 '18 at 12:37
  • @MarkBennet, thanks. I have interpreted it wrong. I thought he subtracted the two inequalities. – gareth618 Jul 17 '18 at 15:22

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