a^2n +b^2n=c^2n
Posted: Tue Nov 06, 2012 6:11 pm
Prove that if $a^{2n}+b^{2n}=c^{2n}$ where $a,b,c$ are positive,then $a^n+b^n>c^n$.Can you do it by geometry?
Use $L^AT_EX$ please, it makes our work a lot easier.Rafe wrote:PROVE THAT IF $a^{2n}+b^{2n}=c^{2n}$ WHERE $a,b,c$ ARE POSITIVE,THEN $a^n+b^n>c^n$. CAN YOU DO IT BY GEOMETRY
Let $a^n=x,b^n=y,c^n=z$. So the equation implies $x^2+y^2=z^2$
So from Pythagoras's theorem, it is possible to construct a right angled triangle with side lengths $x,y,z$ when $z$ is the hypotenuse. Now from triangle inequality, it easily follows $x+y>z$.