Prove that $\sin^7 x + \cos^7 x < 1$ if $0 < x < \frac{\pi}{2}. $

Question

I am not sure how to attack this. Using the Pythagorean identity seems just to make things messier.

Answer 1

As $0<x<\dfrac\pi2,0<\cos x,\sin x<1$

For $0<y<1, y^{a+b}-y^b=y^b(y^a-1)<0\iff y^{a+b}<y^b$ if $a\ge1$

$\implies\sin^7x<\sin^2x,\cos^7x<\cos^2x$