Let $C$ be a constant, $0\leq\gamma\leq1$ and $0<M<1$. Given that $$ C \leq -\gamma x+\frac{x^2}{M-5x},\qquad \forall x\in[0,M/5). $$
I want to give an upper bound of $C$ which is independent of $x$. In particular, I want to show that $$ C\leq -\frac{M\gamma^2}{12} $$ I was trying to minimise the given function above in the domain $[0,M/5)$,but the expression is horrible and not going anywhere close to the answer. Does anyone have any special idea/trick to work on this question?
Indeed the upper bound can go very large in the domain, but that does not matter for us, because C must be at most the minimum point of the function in the domain.
Hint
$$f(x)= -\gamma x+\frac{x^2}{M-5x}\qquad \forall x\in[0,M/5)$$ Let $x=\frac M 5 t$ and, simplifying, consider $$g(t)=\frac{M}{25}\frac{ t\big( (5 \gamma +1)\, t-5 \gamma\big)}{(1-t)}\qquad \forall t\in[0,1)$$ which is simpler. Now $$\frac d {dt}\Bigg(\frac{ t\big( (5 \gamma +1)\, t-5 \gamma\big)}{(1-t)}\Bigg)=-\frac{(1+5\gamma)t^2-2 (1+5\gamma)t+5\gamma} {(1-t) ^2}$$
Solve the quadratic in numerator, select the proper root and you have $t_*$ which is the critical point. Use the second derivative test to check if it is a maximum or a minimum