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?
Best Answer
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.