Prove trigonometric inequality $\sin x\leq 1-\left(\frac{2x}{\pi}-1\right)^2$ I was working on a trigonometric inequality and after some manipulations I needed to prove that:
$$\sin x\leq 1-\left(\dfrac{2x}{\pi}-1\right)^2, \enspace \forall x\in \left[0,\pi\right).$$
My idea was to move the square on a side and then square root and prove what we got. But I failed. Please help me solve this! Thank you! Please don't use calculus for the proof.

 A: Concavity of Sine
For $x,y\in[0,\pi]$,
$$
\begin{align}
\frac{\sin(x)+\sin(y)}2
&=\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)\\[6pt]
&\le\sin\left(\frac{x+y}2\right)\tag1
\end{align}
$$
Since $\sin(x)$ is continuous, $(1)$ shows that $\sin(x)$ is concave on $[0,\pi]$.

The Inequality
Note that for $x=0$ and $x=\frac\pi4$, $\sin(x)=\frac{2\sqrt2x}\pi$. Thus, since $\sin(x)$ is concave on $\left[0,\frac\pi4\right]$, we have
$$
\sin(x)\ge\frac{2\sqrt2x}\pi\tag2
$$
for $x\in\left[0,\frac\pi4\right]$. Thus, for $\frac x2\in\left[0,\frac\pi4\right]$, that is, $x\in\left[0,\frac\pi2\right]$,
$$
\begin{align}
\cos(x)
&=1-2\sin^2\left(\frac x2\right)\tag3\\
&\le1-\frac{4x^2}{\pi^2}\tag4
\end{align}
$$
where step $(4)$ is simply an application of $(2)$. Since $(4)$ is even, it is true for $x\in\left[-\frac\pi2,\frac\pi2\right]$. Therefore, for $\frac\pi2-x\in\left[-\frac\pi2,\frac\pi2\right]$, that is, $x\in[0,\pi]$,
$$
\begin{align}
\sin(x)
&=\cos\left(\frac\pi2-x\right)\tag5\\
&\le1-\frac{4\left(\frac\pi2-x\right)^2}{\pi^2}\tag6\\
&=1-\left(\frac{2x}\pi-1\right)^2\tag7
\end{align}
$$
where $(6)$ is an application of $(4)$.
A: Note that $ f(t)=\frac\pi{\sqrt2}\sin \frac t2,\> t\in [0,\pi/2] $ is a concave function with $f(0)=0$ and $f(\frac\pi2)= \frac\pi2 $, which implies $f(t)\ge t$, i.e.
$$\frac\pi{\sqrt2}\sin \frac t2- t\ge 0
\implies \sin^2\frac t2 \ge (\frac{\sqrt2 t}\pi)^2
\implies 1-\cos t \ge (\frac{2t}\pi)^2  $$
Substitute $x= \frac\pi2+t, \> x\in [0, \pi]$, to obtain
$$\sin x\leq 1-\left(\dfrac{2x}{\pi}-1\right)^2$$
A: We need to prove that $$2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\geq\left(\frac{2x}{\pi}-1\right)^2$$ or
$$\left(\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{2x}{\pi}+1\right)\left(\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\frac{2x}{\pi}-1\right)\geq0.$$
Now, show that $$ \sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{2x}{\pi}+1\geq0$$ and $$\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\frac{2x}{\pi}-1\geq0,$$ which we can prove by using one derivative only:
Let $f(x)=\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{2x}{\pi}+1.$
Thus, $$f'(x)=\frac{1}{\sqrt2}\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)-\frac{2}{\pi}<0,$$
which says $$f(x)>f(\pi)=0.$$
Let $g(x)=\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\frac{2x}{\pi}-1.$
Thus, $$g'(x)=\frac{1}{\sqrt2}\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)+\frac{2}{\pi}>0,$$ which gives $$g(x)\geq g(0)=0$$ and we are done! 
A: I would study the function $ f(x)=\sin x+\frac{4x^2}{\pi^2}-\frac{4x}{\pi}$ should not be a problem to prove that it is negative between $0$ and $\pi$.
