Finding upper bound on x and y

Question

I want to find upper bound on following i equality without calculator. $${x^2}{y^5} \gt {2^x}{5^y} $$

How can I get this done?

Answer 1

It's obvious that $x\neq0$ and $y>0$.

We can rewrite our inequality in the following form. $$\frac{x^2}{2^x}>\frac{5^y}{y^5}.$$

Easy to see that $$\left\{\frac{5^y}{y^5}|y>0\right\}=\left[\left(\frac{e\ln5}{5}\right)^5,+\infty\right)$$ and $$\left\{\frac{x^2}{2^x}|x\neq0\right\}=\mathbb R^{+}$$

Thus, do not exist upper bound for $y$

because for all $y>0$, where $x\neq0$ for which $\frac{x^2}{2^x}>\frac{5^y}{y^5}$ is true.

Now, let $x_0$ is a biggest root of the equation $$\frac{x^2}{2^x}=\left(\frac{e\ln5}{5}\right)^5.$$
Since, $\max\limits_{x>0}\frac{x^2}{2^x}>\left(\frac{e\ln5}{5}\right)^5$ and $\lim\limits_{x\rightarrow+\infty}\frac{x^2}{2^x}=0$,

we see that for all $x<x_0$ there is $y>0$ for which $\frac{x^2}{2^x}>\frac{5^y}{y^5}$ is true.

Thus, $$\sup_{\frac{x^2}{2^x}>\frac{5^y}{y^5} is true}x=x_0=6.252...$$ Done!