<s>[INST] Minimise $x+y$ under the condition $2x+32y \leq 9xy$ where $x,y > 0$ Let $x$ and $y$ be positive real numbers such that $2x+32y\leq 9xy$. Then find the smallest possible value of $x+y$. I tried using AM-GM but it is difficult to find the case of equality. Is Cauchy-Schwarz also helpful? Thanks. 
 [/INST] Rearranging the inequality yields
\begin{align*}
0 &\le 9xy -2x -32y \\
\frac{64}9 &\le 9xy -2x -32y + \frac{64}9 = \bigg(3 x - \frac{32}3\bigg) \bigg(3 y - \frac23\bigg).
\end{align*}
The AM-GM inequality* now gives
$$
\frac83 \le \sqrt{\bigg(3 x - \frac{32}3\bigg) \bigg(3 y - \frac23\bigg)} \le \frac12 \bigg(3 x - \frac{32}3 + 3 y - \frac23 \bigg) = \frac32(x+y) -\frac{17}3,
$$
which implies $x+y \ge 50/9$. Moreover, the case of equality is $3x-32/3 = 3y-2/3$, which has exactly one point of intersection with the line $x+y=50/9$, namely $x=40/9$, $y=10/9$; one confirms that this point is on the boundary $2x+32y=9xy$ as well. In summary, the smallest possible value of $x+y$ is $50/9$.
*AM-GM applies only when the two factors are both positive, so one must check that points with $0\le x\le 32/9$ and $0\le y\le 2/9$ don't satisfy the necessary inequality.
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