What if i just root them !

Algebra Level 2

It is well know that e π > π e e^{\pi} > \pi^{e} . Is it true that e π > π e ? \large{{\color{#3D99F6}\sqrt{e}}^{{\color{#D61F06}\sqrt\pi} }> {\color{#D61F06}\sqrt{\pi}}^{{\color{#3D99F6}\sqrt e}}}\,?

No Yes

Naren Bhandari by Naren Bhandari , 21, India

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2 solutions

Michael Mendrin
Aug 21, 2018

The value of x 1 x x^{\frac{1}{x}} is at maximum where x = e x=e . For any two values a , b a,b , the one closest to e e will have a greater value. π \sqrt{\pi} is closer to e e than e \sqrt{e} , so the right side has the larger value than the left.

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Anirudh Sreekumar
Aug 21, 2018

Consider e x for x ∈ ( 0 , 1 ) e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! ⋯ + x n n ! ⋯ e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! ⋯ + x n n ! ⋯ < 1 + x + x 2 ! + x 3 ! + x 4 ! ⋯ + x n ! ⋯ ( 1 ) x n < x for n > 1 as x ∈ ( 0 , 1 ) 1 + x + x 2 ! + x 3 ! + x 4 ! ⋯ + x n ! ⋯ < 1 + x + x 2 + x 4 + x 8 ⋯ + x 2 ( n − 1 ) ⋯ ( 2 ) n ! ≥ 2 ( n − 1 ) for n ≥ 1 1 + x + x 2 ! + x 3 ! + x 4 ! ⋯ + x n ! ⋯ < 1 + x + x 2 1 − 1 2 Sum of G.P 1 + x + x 2 ! + x 3 ! + x 4 ! ⋯ + x n ! ⋯ < 1 + 2 x ( 3 ) From ( 1 ) , ( 2 ) , ( 3 ) we have, e x < 1 + 2 x for x ∈ ( 0 , 1 ) ( 4 ) Let, t = 1 2 ( π e − 1 ) 0 < t < 1 e t < 1 + 2 t From ( 4 ) ⟹ e 1 2 ( π e − 1 ) < 1 + 2 × 1 2 ( π e − 1 ) ⟹ e 1 2 ( π e − 1 ) < π e ⟹ e π e e < π e ⟹ e π e < π ⟹ e π < π e For x > y > 1 , x a > y a , a > 0 Thus the given statement is FALSE. \begin{aligned}\text{Consider } e^x \text{ for } x\in(0,1)\\ e^x&=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!} \cdots +\dfrac{x^n}{n!}\cdots\\ e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!} \cdots +\dfrac{x^n}{n!}\cdots&<1+x+\dfrac{x}{2!}+\dfrac{x}{3!}+\dfrac{x}{4!} \cdots +\dfrac{x}{n!}\cdots\hspace{4mm}\color{#3D99F6}\small (1) \hspace{4mm} x^n<x \text{ for } n>1 \text{ as } x\in(0,1)\\ 1+x+\dfrac{x}{2!}+\dfrac{x}{3!}+\dfrac{x}{4!} \cdots +\dfrac{x}{n!}\cdots&<1+x+\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{8} \cdots +\dfrac{x}{2^{(n-1)}}\cdots\hspace{4mm}\color{#3D99F6}\small (2)\hspace{4mm} n!\geq2^{(n-1)} \text{ for } n\geq1\\ 1+x+\dfrac{x}{2!}+\dfrac{x}{3!}+\dfrac{x}{4!} \cdots +\dfrac{x}{n!}\cdots&<1+x+\dfrac{\dfrac{x}{2}}{1-\dfrac{1}{2}}\hspace{4mm}\color{#3D99F6}\small \text{Sum of G.P}\\ 1+x+\dfrac{x}{2!}+\dfrac{x}{3!}+\dfrac{x}{4!} \cdots +\dfrac{x}{n!}\cdots&<1+2x \hspace{4mm}\color{#3D99F6}\small (3)\\\\ \text{From } \color{#3D99F6}(1)\color{#333333},\color{#3D99F6}(2)\color{#333333},\color{#3D99F6}(3)\color{#333333}\text{ we have,}\\\\ e^x&<1+2x \hspace{7mm}\text{ for } x\in(0,1)\hspace{4mm}\color{#3D99F6}\small (4)\\\\ \text{Let, } t&=\dfrac{1}{2} \left(\sqrt{\dfrac{\pi}{e}}-1\right)\\ 0&<t<1\\ e^t&<1+2t \hspace{4mm}\color{#3D99F6}\small \text{From} (4)\\\\ \implies e^{\small\tfrac{1}{2} \left(\sqrt{\tfrac{\pi}{e}}-1\right)}&<1+2\times\dfrac{1}{2} \left(\sqrt{\dfrac{\pi}{e}}-1\right)\\ \implies e^{\small\tfrac{1}{2} \left(\sqrt{\tfrac{\pi}{e}}-1\right)}&<\sqrt{\dfrac{\pi}{e}}\\ \implies \dfrac{{\sqrt{e}}^{\small\sqrt{\tfrac{\pi}{e}}}}{\sqrt{e}}&<\sqrt{\dfrac{\pi}{e}}\\ \implies {\sqrt{e}}^{\small\sqrt{\tfrac{\pi}{e}}}&<\sqrt{\pi}\\ \implies {\sqrt{e}}^{\sqrt{\pi}}&<{\sqrt{\pi}}^{\sqrt{e}}\hspace{4mm}\color{#3D99F6}\small \text{For }x>y>1 , \hspace{2mm} x^a>y^a, a>0\\\\ \text{Thus the given statement is }&\color{#D61F06}\text{ FALSE.}\end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Great Solution. I got matched a bit with you. Thank you for the solution. :)

Naren Bhandari - 2 years, 9 months ago

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