prove that $3x - x^3 < 2 \cdot \sin(\pi/2 \cdot x)$ $\forall x \in (0, 1)$ I faced problems of such kind before and usually I would just chech that on the ends of segment values of compared functions are equal and then find their derivatives and make sure that derivative of one function is always greater on some interval or segment. However, in this case it just didn't work. I found derivative but on some interval $A \subset (0, 1)$ one is greater than the other and on another interval $B \subset (0, 1)$ everything is vice versa.
Do you have any other meaningful approaches to this problem?
 A: Let's prove another inequality, first: $\sin$ is a concave function in $[0,\pi]$, so we have (by Jensen's inequality) $$(1-x)\,\sin0+x\,\sin\frac\pi6=\frac{x}2\le\sin\left((1-x)\cdot0+x\cdot\frac\pi6\right)=\sin\frac\pi6x$$ for $x\in[0,1]$, with equality only in the endpoints of the interval. Moreover, both sides of the inequality take values in $[0,1/2]$. In that interval, the function $g(y)=3y-4y^3$ is monotone increasing (derivative $3(1-4y^2)\ge0$), and the well-known triplication formula says $$g(\sin\phi)=\sin3\phi,$$ so we have
$$g(x/2)\le\sin\frac\pi2x.$$ Multiplying both sides by $2$, we arrive at $$3x-x^3\le2\sin\frac\pi2x$$ on $[0,1]$, and as it's clear from the derivation, there's equality only at the endpoints of the interval.
A: Let $$
f(x) = 2 \cdot \sin(\pi/2 \cdot x) - 3x - x^3 $$
So you want to prove $f(x) \ge 0$ on $[0 \quad 1]$. Now $f(0) = f(1) = 0$ so for all values within the interval you want $f(x) > 0$.
By expansions, check the behavior of $f(x)$ in the vicinity of the interval boundaries. Near $x=0$, you have $f(x) \simeq  (\pi -3)x \simeq 0.14 x$, and near $x=1$, you have $f(x) \simeq  (6 - \pi^2/2) (x-1)^2 \simeq 1.07 (x-1)^2$, so indeed $f(x)$ is rising on both interval boundaries.
Now what is left to show is that $f(x)$ will not fall again below zero within the interval. This is achieved if $f(x)$ indeed only has one single extremal point in $(0,1)$. This can be checked by computing   $f'(x) = \pi \cos((\pi x)/2) + 3 x^2 - 3$. A standard curve discussion (below) gives the result that there is indeed only one solution  $f'(x) =0$ in $(0,1)$. We are done.
Remark: "Simple" Taylor expansion won't work since $f(0) = f(1) = 0$.

Curve discussion. The curve discussion is standard but a bit lengthy since we start identifing a derivative which doesn't change sign in $(0,1)$. With that information, and with the values of all the derivatives at the interval boundaries, we can argue backwards. We need to find the relevant maxima and minima only approximately, since all we want to establish is that  there is only one solution  $f'(x) =0$ in $(0,1)$, and we are not interested in its position.
We have
$$
f'(x) = \pi \cos((\pi x)/2) + 3 x^2 - 3\\
f''(x) =  6x - (\pi^2 \sin((\pi x)/2))/2 \\
f^{(3)}(x) = 6 - (\pi^3 \cos((\pi x)/2))/4 \\
f^{(4)}(x) = (\pi^4 \sin((\pi x)/2))/8
$$
Now we can argue backwards: we have in $(0,1)$ that $f^{(4)}(x) > 0$.
Therefore $f^{(3)}(x)$ is rising strictly monotonously from -1.75 to 6. So in  $(0,1)$, we have just one $x^*$ where $f^{(3)}(x^*) = 0$ which is roughly at  $x^* = 0.4365$.
This means that $f''(x)$, starting from $f''(0) = 0$, is first falling until $x^* = 0.4365$, then again rising to  $f''(1) = 1.0652$. This again means that in $(x^*, 1)$, there is just one  $x^{**}$ where $f''(x^{**}) = 0$ which is roughly at  $x^{**} = 0.77$.
Now we can describe the behavior of $f'(x)$, starting from $f'(0) = 0.1416$, is falling until  $f'(x^{**}) = -0.11$, and then rising until $f'(1) = 0$. Hence there is only one value $x^{***}$ where $f'(x^{***}) = 0$,  where $0 < x^{***} < x^{**}$.
