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

[anemone]

Prove that $6^{33}>3^{33}+4^{33}+5^{33}$.

 

[kaliprasad]

Prove that $6^{33}>3^{33}+4^{33}+5^{33}$.

Spoiler

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)

 

[chisigma]

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

$\chi$ $\sigma$

 

[anemone]

Thank you both for participating and providing us the neat and elegant proof! Well done!(Sun)