5. Zadatak

Prvo rješenje. Imamo

$$\begin{aligned}\sin^4 x+\cos^4 x&=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\\&=1-\frac12\sin^2 2x\geq1-\frac12=\frac12.\end{aligned}$$

Drugo rješenje. Slično dobivamo

$$\begin{aligned}\sin^4 x+\cos^4 x&=(\sin^2 x)^2+(1-\sin^2 x)^2=2\sin^4 x-2\sin^2 x+1\\&=\frac12(4\sin^4 x-4\sin^2 x+1)+\frac12\\&=\frac12(2\sin^2 x-1)^2+\frac12\geq\frac12.\end{aligned}$$

Treće rješenje. Polazna nejednakost je ekvivalentna sljedećima:

$$\begin{aligned}\sin^4 x+\cos^4 x&\geq\frac12\\\iff(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x&\geq\frac12\\\iff1-2\sin^2 x\cos^2 x&\geq\frac12\\\iff4\sin^2 x\cos^2 x&\leq1\\\iff\sin^2 2x&\leq1,\end{aligned}$$

a ovo uvijek vrijedi. Zato vrijedi i polazna nejednakost.