inequality involving $x$, $x^3$,$\sin(x),\cos(x)$

Question

Let $x \in \left[0,\dfrac {\pi} 2 \right]$. Prove the inequality

$$6x \ge 6\sin x +x^3 \cdot \cos x$$

there is nice solution using Taylor expansion. Is there other one?

Answer 1

Over the interval $I=\left(0,\frac{\pi}{2}\right)$ we have: $$\frac{\tan x}{x}\geq 1 \geq \frac{1-\frac{x^2}{6}}{1+x^2}\tag{1}$$ so the previous inequality gives that: $$ \frac{d^2}{dx^2}\left(6\sin x+x^3\,\cos x\right)=(6x-x^3)\cos x-(6+6x^2)\sin x\leq 0\tag{2}$$ over $I$, so $f(x)=6\sin x+x^3\cos x$ is a concave function on $I$ and the claim follows by computing the tangent line through the origin, $g(x)=6x$.

Answer 2

(EDIT) Let $f(x) = 6 x - 6 \sin(x) - x^3 \cos(x)$. $f''(x) = 6 (\sin(x) - x \cos(x)) + 6 x^2 \sin(x) + x^3 \cos(x)$.
Now $\sin(x) - x \cos(x) = \cos(x) (\tan(x) - x) \ge 0$ for $x \in (0,\pi/2)$ while $x^2 \sin(x) \ge 0$ and $x^3 \cos(x) \ge 0$ in this interval as well, so $f''(x) \ge 0$ for $x \in [0,\pi/2]$.

But $f(0)=0$ and $f'(0) = 0$, so $$f(x) = f(0) + f'(0) x + \int_0^x (x-t) f''(t)\; dt \ge 0$$ for $0 \le x \le \pi/2$.