Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number. Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number.

Slightly painful way (there may be more elegant):
Since $\sin^2+\cos^2 = 1$, you can rewrite this as
$$
(1-\cos^2 x)^4 + \cos^8 x \geq \frac{1}{8}
$$
or equivalently 
$$
2 \cos^8 x - 4\cos^6 x + 6\cos^4 x - 4\cos^2 x + 1 \geq \frac{1}{8}.
$$
Setting $X=\cos^2 x$, you want first to solve the inequality
$$
2 X^4 - 4X^3 + 6X^2 - 4X + \frac{7}{8} \geq 0
$$
i.e., once factored:
$$
\frac{1}{8}(2X-1)^2(4X^2-4X+7) \geq 0
$$
but the polynomial $4x^2-4x+7$ has no real roots—compute the discriminant) and is always positive; so since the other factors are clearly non-negative, the inequality is true for all $X$. It follows (going back to $\cos x$) that the original inequality also holds for all $x$. 
