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?
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Do you want an upper bound of x in terms of y (and vice versa), or do you want an upper bound for x that holds for all y (and vice versa)? – ZTaylor Jul 01 '17 at 07:21
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an upper bound for x that holds for all y – Muhammad Saboor Jul 01 '17 at 07:22
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If all $y$, $y=-1$ leads false. – Takahiro Waki Jul 01 '17 at 07:50
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lower bound is x,y > 0 – Muhammad Saboor Jul 01 '17 at 07:59
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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!
Michael Rozenberg
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how can we say that for all x > 0 there is y > 0 for which inequality is true. – Muhammad Saboor Jul 01 '17 at 07:27
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@Muhammad Saboor I made a mistake. I fixed my post. See now please. – Michael Rozenberg Jul 01 '17 at 08:04
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can you tell me how $$ \left{\frac{5^y}{y^5}|y>0\right}=\left[\left(\frac{e\ln5}{5}\right)^5,+\infty\right) $$ – Muhammad Saboor Jul 01 '17 at 08:43
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@Muhammad Saboor Because $\lim\limits_{y\rightarrow+\infty}\frac{5^y}{y^5}=+\infty$. $\left(\frac{5^y}{y^5}\right)'=\frac{5^y\ln5}{y^5}-\frac{5\cdot 5^y}{y^6}=\frac{5^y(y\ln5-5)}{y^6}$, which gives $y_{min}=\frac{5}{\ln5}$ and $\frac{5^y}{y^5}\geq\frac{5^{\frac{5}{\ln5}}}{\left(\frac{5}{\ln5}\right)^5}=\left(\frac{5^{\frac{1}{\ln5}}\ln5}{5}\right)^5=\left(\frac{e\ln5}{5}\right)^5.$ – Michael Rozenberg Jul 01 '17 at 09:06