P&B #106

This is a problem from Putnam and Beyond:

Given that $9a^2 + 8ab + 7b^2 \le 6$, show that $7a + 5b + 12ab \le 9$.

My solution is extremely janky; nevertheless, it is still a solution that is slightly different from the book’s.

Our hypothesis implies that $-12ab \ge -9 + \frac{27}{2}a^2 + \frac{21}{2}b^2$ by manipulation. Note that our conclusion is equivalent to $9 – 7a – 5b – 12ab \ge 0$. Hence, we have

\begin{align*}
9 – 7a – 5b – 12ab &\ge \frac{27}{2}a^2 + \frac{21}{2}b^2- 7a – 5b \stackrel{?}{\ge} 0.
\end{align*}

Now consider the two quadratics $\frac{27}{2}a^2 – 7a$ and $\frac{21}{2}b^2 – 5b$. They are negative between $0$ and their other root, which is $a = \frac{14}{27}$ and $b = \frac{10}{21}$. Since the objective function is strictly increasing for in the positive numbers, checking the last root reveals that it is indeed satisfied as $7a + 5b + 12 ab = \frac{1696}{189} < 9$. Again, janky but doable in a reasonable time. (Of course, there's more proper methods with Lagrange multipliers and such...)

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