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. 
 A: 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.
A: If you want an approach with CS inequality:
$$(y+x)(x+16y)\ge (\sqrt{xy}+4\sqrt{xy})^2 \implies x+y \ge 2\frac{25xy}{2x+32y}\ge \frac{50}9$$
Equality is when $\dfrac{x}{y}=\dfrac{16y}{x} \iff x=4y$ and $2x+32y=9xy$, together giving $x = \dfrac{40}9, \; y = \dfrac{10}9$.
A: We can approach the problem geometrically. $2x + 32y = 9xy$ is a hyperbola, dividing the plane in three regions, we can easily work out which regions satisfy the inequality. Now consider the set of lines $x + y = t$ where $t$ is a constant. As $t$ varies, the line $x + y = t$ slides left-right, we want the least positive $t$ that the line $x + y = t$ intersects the region satisfying the inequality. By the symmetry of the hyperbola, we can do this by solving the simultaneous equations:
$$2x + 32y = 9xy$$
$$x + y = t$$
To find the value of $t$ that give only one solution in $(x,y)$.
A: here is how i approached. i will look at the boundary of $$2x + 32 y = 9xy  $$  which is given by the hyperbola $$ y = \frac{2x}{9x - 32}, \frac{dy}{dx} = -\frac{64}{(9x-32)^2} $$ this hyperbola has slope of $-1$ at $$ -\frac{64}{(9x-32)^2} = -1 \to \left(x = \frac{40}9,y = \frac{10}9, x + y = \frac{50} 9\right) \text{ and} \left(x = \frac{24}9,y = -\frac{6}9, x + y = 2\right) $$ the second solution must be rejected because of the condition that $x > 0$ and $y > 0.$
