Prove (6/5)^√3 > (5/4)^√2

[anemone]

Hi MHB!

I am back to posting our High School POTW.(Smile) During my hiatus, MarkFL has been so supportive and his willingness to fill in for me is greatly appreciated.(Yes)(Beer)

Here is this week's POTW:

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Prove that $\left(\dfrac{6}{5}\right)^{\sqrt{3}}>\left(\dfrac{5}{4}\right)^{\sqrt{2}}$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week POTW. :( You can find the suggested solution as follows:

Spoiler

For $0<x<1$, we have $\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}-\cdots$.

$\therefore x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}>\ln(1+x)>x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}$

Substituting $x=\dfrac{1}{5}$ and $x=\dfrac{1}{4}$ into the inequalities above we get

$\ln\dfrac{6}{5}>\dfrac{1367}{7500}$ and $\dfrac{857}{3840}>\ln \dfrac{5}{4}$

If we can prove $\sqrt{3}\dfrac{1367}{7500}>\sqrt{2}\dfrac{857}{3840}$, then we are done.

But that is simply true since

$\sqrt{3}\times 1368 \times 3840 <\sqrt{2} \times 857 \times 7500$

$\sqrt{3}\times 1367 \times 64 >\sqrt{2}\times 857 \times 125$

$22962450432>22951531250$

$\therefore \left(\dfrac{6}{5}\right)^{\sqrt{3}}>\left(\dfrac{5}{4}\right)^{\sqrt{2}}$ (Q.E.D.)