If $a,b,c$ are sides of right triangle with $c$ hypothenuse then show that for every positive integer $n>2$ we have $c^n>a^n+b^n$ .
Problem
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Tags: inequalities
tarzanjunior
09.01.2017 19:05
Use Power mean inequality
SAUDITYA
10.01.2017 18:06
$c^2 = a^2 + b^2 => c^n = c^{n-2}a^2 + c^{n-2}b^2 > a^n + b^n$.
adityaguharoy
10.01.2017 18:10
SAUDITYA wrote: $c^2 = a^2 + b^2 => c^n = c^{n-2}a^2 + c^{n-2}b^2 > a^n + b^n$. is the same as that meant by : tarzanjunior wrote: Use Power mean inequality
Rijadinho
02.12.2023 13:08
$c^2=a^2+b^2$ $c^n=c^{2}c^{n-2}=a^{2}c^{n-2}+b^{2}c^{n-2}$ $c>a,b \Rightarrow c^{n-2}>a^{n-2} \wedge c^{n-2}>b^{n-2}$ $c^n=a^{2}c^{n-2}+b^{2}c^{n-2}>a^{2}a^{n-2}+b^{2}b^{n-2}=a^n+b^n$ $\blacksquare$