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.
 A: Hint:apply the inequality thrice: $x^2+y^2 \geq \dfrac{(x+y)^2}{2}$.
A: $$\sin^8 x+\cos^8x \ge \frac 18;$$
$$ \left (\frac{1-\cos2x}{2} \right )^4+ \left (\frac{1+\cos2x}{2} \right )^4\ge \frac 18;$$
$$(1-\cos2x)^4+(1+\cos2x)^4 \ge2$$
$$1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x+$$
$$+1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x\ge 2$$
$$12\cos^22x+2\cos^42x \ge 0$$
A: 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 $x$. 
A: By derivation, the extrema are reached for
$$\cos(x)\sin^7(x)-\sin(x)\cos^7(x)=0=\cos(x)\sin(x)(\sin^6(x)-\cos^6(x)).$$
They occur at multiples of $\frac\pi4$, where the function equals $0+1$ or $\frac1{16}+\frac1{16}$ or $1+0$.
