MHB Can You Prove $6^{33}>3^{33}+4^{33}+5^{33}?

AI Thread Summary
The discussion centers on proving the inequality $6^{33} > 3^{33} + 4^{33} + 5^{33}$. Participants utilize the binomial expansion to demonstrate that the terms derived from $(1+x)^{33}$ for specific values of x lead to inequalities that support the original claim. Specifically, they show that contributions from the terms involving $5^{33}$ exceed those from $4^{33}$ and $3^{33}$. The proof is confirmed to be valid, affirming the inequality. The conversation concludes with appreciation for the elegant proof provided by the contributors.
anemone
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Prove that $6^{33}>3^{33}+4^{33}+5^{33}$.
 
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anemone said:
Prove that $6^{33}>3^{33}+4^{33}+5^{33}$.

We know $6^3 = 3^3 + 4^3 + 5^3$ as both sides are 216
Multiply by $6^{30}$ on both sides

$6^{33} = 6^{30}( 3^3 + 4^3 + 5^3)$
$= 6^{30}* 3^3 + 6^{30} * 4^3 + 6^{30} * 5^3$
$> 3^{30} * 3^3 + 4^{30} * 4^3 + 5 ^ {30} * 5^ 3$
$> 3^{33} + 4^{33} + 5^{ 33}$

as a matter of fact $6^n > 3^n + 4^n + 5^n$ for n > 3 (not even integer)
 
anemone said:
Prove that $6^{33}>3^{33}+4^{33}+5^{33}$.

[sp]Is...

$\displaystyle (1+x)^{33} = x^{33} + 33\ x^{32} + 33\ \cdot\ 16\ \cdot\ x^{31} + ...\ (1)$

... and for x=5...

$\displaystyle (1+5)^{33} = 5^{33} + 33\ 5^{32} + 33\ \cdot\ 16\ \cdot\ 5^{31} + ...\ (2)$

But...

$\displaystyle 33\ 5^{32} = \frac{33}{5}\ 5^{33} = \frac{33}{5}\ (\frac{5}{4})^{33}\ 4^{33} > 4^{33}\ (3)$

... and...

$\displaystyle 33\ \cdot 16\ \cdot 5^{31} = \frac{33}{25}\ \cdot 16\ \cdot 5^{33} = \frac{33}{25}\ \cdot 16\ \cdot (\frac{5}{3})^{33}\ \cdot 3^{33} > 3^{33}\ (4) $

... so that Your assumption is true...[/sp]

Kind regards

$\chi$ $\sigma$
 
Thank you both for participating and providing us the neat and elegant proof!:cool: Well done!(Sun)
 
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