If we assume that $a,b$ are real numbers such that $9a^2+8ab+7b^2\le 6$, how to prove that :
$7a+5b+12ab\le9$
If we assume that $a,b$ are real numbers such that $9a^2+8ab+7b^2\le 6$, how to prove that :
$7a+5b+12ab\le9$
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.