Minimising the upper bound of a function

Let $$\gamma\in[0,1]$$, $$M>0$$, $$n\in\mathbb{Z}^{+}$$ and $$P\leq \exp\left(\frac{M}{5}\right)\exp\left(-n\gamma x + \frac{nx^2}{M-5x}\right),\qquad \text{for }x\in[0, M/5).$$ I want to show that I can reformulate an upper bound that is independent of $$x$$, by minimising it in $$[0,M/5)$$. In particular, I want to show that $$P\leq \exp\left(\frac{M}{5}\right)\exp\left(-n\frac{M\gamma^2}{12}\right).$$ $$P$$ denotes a probability and can only output values between $$0$$ and $$1$$. $$P$$ is independent of $$x$$.

This new upper bound works when I check it using some graphing tools. However, I want to derive it analytically from the info I have here. Also, the value $$\exp(-nM\gamma^2/12)$$ is not the local minimum point of the original bound in $$[0,M/5)$$. It is just a value large enough for us to justify this uniform bound. Can anyone provide any insights to this question?

This doesnt seem right since for $$x = M/5-\epsilon$$ the RHS is arbitrarily large.
• We still have the same issue. Are you sure you are not looking to find that $P$ is bounded above by the expression you showed for some interval around $0$? if you include values arbitrarily close to $x=M/5$ then the upper bound is unbounded (and thus trivial)