TITLE: Show that the function $f(x,y)=\frac{\frac1x+y}{\frac1{x(x+1)}+y+\frac1y}$ is minimized at $x=y=1$
QUESTION [3 upvotes]: Define the function
$$f(x,y)=\frac{\frac1x+y}{\frac1{x(x+1)}+y+\frac1y}$$
for $0<x\leq 1$ and $1\leq y\leq 2$. Show that the function is minimized at $x=y=1$.

While the function is decreasing in $x$ for $y=1$, it is not necessarily decreasing in $x$ for other fixed $y$. For example, $f(0.1,2)=1.035...$, while $f(0.2,2)=1.05$.

REPLY [1 votes]: I have used Mathematica to compute
$$f_y={x(1+x)\bigl(1+x+xy(2+2x-y)\bigr)\over\bigl(\cdots\bigr)^2}\ .$$
This shows that $f_y\geq0$ in the domain at stake, and implies that it is sufficient to minimize
$$g(x):=f(x,1)\qquad(0<x\leq1)\ .$$
Mathematica then produced
$$g'(x)=-{2x(1+x)\over\bigl(\cdots\bigr)^2}<0\qquad(0<x\leq1)\ ,$$
so that we can safely say that $f$ takes its minimum in $\>]0,1]\times[1,2]$ at the point $(1,1)$, with value $f(1,1)={4\over5}$.
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