Prove $7^{71}>75^{32}$ My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$
I did the first question: \begin{align}\frac{6^9}{10^7}&=\frac{4}{5}\times\frac{27^3}{25^3}\\&=0.8\times1.08^3\\&>0.8\times(1+3\times0.08+3\times0.08^2)\\&>0.8\times(1+3\times0.086)\\&>0.8\times1.25=1.\end{align}
But I can't work out the second,  I calculated it out on my computer, $\frac{7^{71}}{75^{32}}=1.000000949\cdots$
 A: Just for your curiosity :
Using Young's inequality for product we get for $x>1$ :
$$f(x)=\left(\frac{(\frac{75}{7})^{x}}{x}+\frac{x-1}{x}7^{\frac{x}{x-1}}\right)>75$$
Now we have to show for some $x>1$ :
$$g(x)=\frac{f(x)}{49*7^{\frac{7}{32}}}<1$$
And now the miracle of the inequality is :
$$g\Big(\frac{71}{39}\Big)<1$$
And as we now $71=39+32$
So the inequality can be rewritten :
$$\frac{\ln(7)}{\ln(75)-\ln(7)}+1>\frac{71}{39}$$
The LHS is the equality case for the Young's inequality .
So now I suspect we can use Reductio ad absurdum to solve it .
Well some explanations :
If we suppose :
$$7^{71}<75^{32}$$
It's easy to remark that it's equivalent to :
$$\frac{75*39}{71}+\frac{32}{71}7^{\frac{71}{32}}>75$$
By the hypothesis wich make the reductio ad absurdum plausible we have :
$$\Big(\frac{75}{7}\Big)^{\frac{71}{39}}>75$$
So we have :
$$\frac{\Big(\frac{75}{7}\Big)^{\frac{71}{39}}39}{71}+\frac{32}{71}7^{\frac{71}{32}}>\frac{75*39}{71}+\frac{32}{71}7^{\frac{71}{32}}>75$$
Wich is the Young's inequality !
Now to get the end we need to show :
$$7^{\frac{71}{32}}>\frac{\Big(\frac{75}{7}\Big)^{\frac{71}{39}}39}{71}+\frac{32}{71}7^{\frac{71}{32}}$$
So we need to find an upper bound instead of a lower bound .
Any idea ?
