how to find the maxima of $\sin^3x+\cos^3x$ without using derivative

Question

With derivative ,it's easy to do.

$$3 \sin ^2(x) \cos (x)-3 \sin (x) \cos ^2(x)=0$$

$x=0,\max=1$

but without it,how?

Answer 1

$\sin^3 (x)\leq \sin^2(x)$ and $\cos^3(x)\leq \cos^2 x$

$\Longrightarrow \sin^3(x)+\cos^3(x)\leq \sin^2(x)+\cos^2(x)=1.$

Equality hold when $x=2n\pi$ or $\displaystyle x=(4n+1)\frac{\pi}{2}$

Answer 2

We know that $\sin(x)\leq 1$ and $\cos(x)\leq 1$, which means that $$ \sin^3(x)\leq \sin^2(x)\\ \cos^3(x)\leq \cos^2(x) $$ with equality exactly when $x$ is of the form $4n\frac\pi2$ or $(4n + 1)\frac\pi2$ for integer $n$ (when one of the inequalities reads $0\leq 0$ and the other one reads $1\leq 1$). This gives $$ \sin^3(x) + \cos^3(x)\leq \sin^2(x) + \cos^2(x) = 1 $$ with equality exactly when $x$ is on the form $4n\frac\pi2$ or $(4n + 1)\frac\pi2$ for integer $n$.