Trig question

Oh no! My queen NotThatSmart's answer has been blocked by the forces of evil! Maybe she isn't smart enough??!!! HAHA IM JUST KIDDING GURLLL!!!!!

By the way, please fix that ATROCIOUS SLOP on the second to last line so your answer can be an absolute EAT. 

SALAYYYY QUEEN!!!

This problem is a bit tricky, but it's solvable using some handy equations. 

First, let's rewrite the inequality. We have

\( \sin^7(a) + \cos^7(a) < 2 \sin^4(a) \cos^4(a) \)

Now, let's let \(x = \sin(a)\) and \( y = \cos(a)\)

Since we have \( 0 < a < (\pi)/(2)\) and  \( 0 < x, y < 1 \), we can write the inequality

\( x^7 + y^7 < 2 x^4 y^4 \)

Now, we can apply the Arithmetic Mean-Geometric Mean Inequality or the AM-GM to x^4 and y^4, we have

\( (x^4 + y^4)/(2) \geq \sqrt{(x^4 y^4) }\\ x^4 + y^4 \geq 2 x^2 y^2 \)

\( x^4 y^4 \leq \left( (x^4 + y^4)/(2) \right)^2\)

Now, also note that

\( x^7 + y^7 < 2 x^4 y^4 \) since x and y have to be between 0 and 1. 

Now, we can prove this same thing with sin(a) and cos(a). 

We have \( 0 < a < (\pi)/(2)\) and the fact that both sin(a) and cos(a) are between 0 and 1. 

Now, using the same logic above, we can clearly see that 

\( \left(\sin(a)\right)^7 + \left(\cos ( a)\right)^7< 2*sin(a)^4*cos(a)^4\)

Thanks! :)