Optimal values for $\frac{5-2f(x)-3f(y)}{5-g(x)-2g(y)}$ Let $f,g$ be nondecreasing functions from $[0,1]\rightarrow[0,1]$ with $f(0)=g(0)=0$ and $f(1)=g(1)=1$. Let $X=\{(x_0,y_0)\in[0,1]\}$ be the set of values maximizing the function $$h(x,y)=\frac{5-2f(x)-3f(y)}{5-g(x)-2g(y)}.$$ Is it always true that for some $(x_0,y_0)\in X$, $x_0\leq y_0$?
For example, if $f=g$, then the numerator is always no more than the denominator, so we should take $x_0=y_0=0$.
If $f(x)=x^2$ and $g(x)=x$, then $x_0\approx 0.28$ and $y_0\approx 0.37$, so $x_0<y_0$.
 A: It is not true. Take functions
$$f(x) = g(x) = \left\{
\begin{array}{r l}
0, & x\in [0,\frac 12]\\
2x - 1,& x\in [\frac 12, 1]
\end{array}\right. $$
then $$h(x,y) = \frac{5-2f(x)-3f(y)}{5-f(x)-2f(y)} = 1 - \frac{f(x)+f(y)}{5-f(x) -2f(y)}\leq 1$$ and $X = \{(x,y)\,|\, f(x)= f(y) = 0\}$. Let $x=\frac 1 2$ and $y = 0$ for your counterexample.
Note that particular choice of $f$ is irrelevant, whenever $f=g$ we have $X$ as above and $(0,0)\in X$, just as you noted. The problem is that $|f^{-1}(\{0\})|>1$, so the question could still be valid if $f$ and $g$ were strictly increasing.
Actually, whenever $f\geq g$, we have that $$h(x,y) \leq \frac{5-2g(x)-3g(y)}{5-g(x)-2g(y)} = 1 - \frac{g(x)+g(y)}{5-g(x) -2g(y)}\leq 1$$
Noting that $h(0,0) = 1$, the above inequality implies that $(x,y)\in X\implies g(x) = g(y) = 0$. Now, if $h(x,y) = 1$ and $g(x) = g(y) = 0$, we have that $$\frac{5-2f(x)-3f(y)}{5} = 1\implies 2f(x) + 3f(y) = 0 \implies f(x) = f(y) = 0$$
and, because $f\geq g$, $$f(x) = f(y) = 0 \implies g(x) = g(y) = 0,$$ so, $X = \{(x,y)\,|\, f(x) = f(y) = 0\}$. This gives many counterexamples if $|f^{-1}(\{0\}) |>1$.
