<s>[INST] Minimizing the expression $(1+1/x)(1+m/y)$ over positive reals such that $mx+y=1$ Let $x$ and $y$ be positive real numbers such that $mx+y=1$. Find the positive $m$ such that the minimum of: 
$$\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right).$$
is $81$. 
I have tried expanding, using Cauchy-Schwarz inequality, but nothing helped.
 [/INST] Since $x=\frac{1}{m}(1-y)$, we have:
$$\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=\left(1+\frac{m}{1-y}\right)\left(1+\frac{m}{y}\right),\tag{1}$$
and since $f(x)=1+\frac{m}{x}$ is a log-convex function, the minimum of the RHS of $(1)$ over $(0,1)$ occurs when $y=\frac{1}{2}$, and the value of such a minimum is just $(1+2m)^2$. The answer is so $m=4$.

In order to avoid log-convexity, we can just write down the RHS of $(1)$ in the following form:
$$ g(y) = 1+\frac{m+m^2}{y(1-y)}$$
and notice that $y(1-y)\leq \frac{1}{4}$ due to the AM-GM inequality, from which:
$$ f(y) \geq 1+4m+4m^2 = (2m+1)^2 $$
with equality occurring only when $y=(1-y)$, i.e. $y=\frac{1}{2}$.
</s>