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)

 

[CellCoree]

I cannot "prove" mathematical equations, as that is the realm of mathematicians. However, I can use mathematical principles and reasoning to support the statement that $6^{33}>3^{33}+4^{33}+5^{33}$.

Firstly, we can simplify the equation to $6^{33}>3^{33}+4^{33}+5^{33}$ by dividing both sides by $3^{33}$. This results in $6^{33}>1+4^{33}+5^{33}$.

Next, we can use the principle of exponents, which states that for any real numbers $a$ and $b$, if $a>b$, then $a^n>b^n$ for any positive integer $n$. Since $6>5$, we can raise both sides of the equation to the 33rd power, resulting in $6^{33}>5^{33}$.

Combining this with our previous simplification, we get $6^{33}>1+4^{33}+5^{33}>5^{33}$.

Furthermore, using the principle of inequalities, we know that adding a positive number to an already larger number will result in an even larger number. In this case, adding $4^{33}$ to $5^{33}$ will result in a number larger than $5^{33}$.

Therefore, we can conclude that $6^{33}>5^{33}>1+4^{33}+5^{33}$, and thus, $6^{33}>3^{33}+4^{33}+5^{33}$. While this may not be a "proof" in the strict mathematical sense, it provides logical reasoning and evidence to support the statement that $6^{33}$ is indeed greater than the sum of $3^{33}$, $4^{33}$, and $5^{33}$.