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