Let [tex]y=\cos^2x[/tex]. Then
[tex]\cos^8x+\sin^8x=\cos^8x+(1-\cos^2x)^4=y^4+(1-y)^4[/tex]
If calculus methods are okay, consider the function [tex]f(y)=y^4+(1-y)^4[/tex]. Then [tex]f[/tex] has a critical point where [tex]f'[/tex] vanishes, i.e.
[tex]f'(y)=4y^3-4(1-y)^3=0\implies y^3=(1-y)^3\implies y=\dfrac12[/tex]
The second derivative [tex]f''[/tex] is
[tex]f''(y)=12y^2+12(1-y)^2[/tex]
and at this critical point, we have a value of
[tex]f''\left(\dfrac12\right)=6>0[/tex]. The derivative test for extrema indicates this is the site of a minimum, and we get a minimum value of
[tex]f\left(\dfrac12\right)=\dfrac18[/tex]
as desired.
