<s>[INST] inequality involving $x$, $x^3$,$\sin(x),\cos(x)$ 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?
 [/INST] (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 $0 \le x \le \pi/2$.
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