<s>[INST] Inequality: $7a+5b+12ab\le9$ If we assume that $a,b$ are real numbers such that $9a^2+8ab+7b^2\le 6$, how to prove that :
$$7a+5b+12ab\le9$$
 [/INST] We have 
$$2(a-b)^2+7\left(a-\frac{1}{2}\right)^2 + 5\left(b-\frac{1}{2}\right)^2 \geq 0$$
which is equivalent to
$$7a+5b+12ab\leq 9a^2+7b^2+8ab+3 \leq 6+3=9$$
The motivation here is to search for equality case by solving the system of equation in real values $a,b$
\begin{equation*}
\begin{cases}
7a+5b+12ab=9 \\
9a^2+7b^2+8ab=6
\end{cases}
\end{equation*}
which yields $a=b=\frac{1}{2}$. Thus the factors $\left(a-\frac{1}{2}\right)^2$, $\left(b-\frac{1}{2}\right)^2$ and $(a-b)^2$ are in order.
</s>