Proof that $\text{argmax}_x (f_\text{new}(x)) < \text{argmax}_x (f_\text{old}(x)) \text{ if } c_\text{new}(x) > c_\text{old}(x)$ for all $x$ I want to prove the following: Suppose we have the following function $f(x) = y(x)  \cdot (m-c(x)) $ and we know that $y(x)$ and $c(x)$ are positive, concave, and increasing.
I want to prove that when I change $c(x)$ such that $c_\text{new}(x) > c_\text{old}(x)\; \forall x$, that $\text{argmax}_x (f_\text{new}(x)) < \text{argmax}_x (f_\text{old}(x))$
Intuitively, this makes sense to me: We multiply $y(x)$ (increasing function) with the negative of $c(x)$, where -- for lack of more formal words -- $c_\text{new}$ constitutes a larger  "penalty" than $c_\text{old}$.
How would you approach a problem such as this? Thank you!
 A: I believe I found a counterexample on the domain $(0,1)$:

*

*Let $m=1$

*Let $y(x)=\sqrt x$

*Let $c_{old}(x)=\sqrt x$

*Let $c_{new}(x)=\frac{1+x}{2}$
Then $f_{old}(x)=\sqrt x(1-\sqrt x)=\sqrt x-x$ with maximum at $f'_{old}(x)=0$ which gives $\frac{1}{2\sqrt x}-1=0$, i.e. $x=\frac14$.
But $f_{new}(x)=\sqrt x\left(1-\frac{1+x}{2}\right)=\frac{1-x}{2}\sqrt x$ has maximum at $f'_{new}(x)=0$ which gives $\frac{1}{4\sqrt x}-\frac{3\sqrt x}{4}=0$, i.e. $x=\frac 13 > \frac 14$.
If instead, you wanted $\mathrm{argmax}_x(f_{new})>\mathrm{argmax}_x(f_{old})$, a counterexample would be given by $c_{new}(x)=2\sqrt x$, which gives $f_{new}'(x)=0\Rightarrow \frac{1}{2\sqrt x}-2=0\Rightarrow x=\frac{1}{16}$.
EDIT:
Of course, I should show that $c_{new}\geq c_{old}$, but this is equivalent to $1+x\geq2\sqrt x$, which is equivalent to $(1-\sqrt x)^2\geq0$, which is clearly true.
EDIT 2: (copied from comments)
When you want $c_{new}(0)=c_{old}(0)$ you can take the slightly modified $$c_{new}(x)=\frac{101}{200}(x^{0.01}+x)$$ instead, and all others ($m,y,c_{old}$) the same. You can numerically check that this solution works as well.
