Showing $\cos^{10}x\le 1-x^2$ for $x$ in $[0,0.5]$ I have to prove this inequality: 

$$\cos^{10}x\le 1-x^2 \quad \text{for all}\quad x \in [0,0.5]$$ 

Is there any easy and/or elegant way to do this? I can do this with Taylor, but it's really a mess. :(
Thanks in advance
 A: Note that
$$\cos(x)^{10}\le 1-x^2\iff 10\log\cos x\le\log (1-x^2)$$
and


*

*$10\log\cos x\le 10 \log
   (1-\frac12x^2+\frac1{24}x^4)\le-5x^2+\frac5{12}x^4\quad$ by $\log(1-x)<x$

*$\log (1-x^2)\ge-x^2-x^4$ to be proved


and since
$$-5x^2+\frac5{12}x^4\le -x^2-x^4 \iff4x^2-\frac{17}{12}x^4\ge 0\iff x^2(4-\frac{17}{12} x^2)\ge 0\\\iff -\sqrt{\frac{48}{17}}\le x\le \sqrt{\frac{48}{17}}$$
the inequality is proved.
To prove 


*

*$\log (1-x^2)\ge-x^2-x^4$


let consider


*

*$g(x)=\log (1-x^2)\ge+x^2+x^4$


and note that


*

*$g(0)=0$

*$g'(x)=\frac{2x^3-4x^5}{1-x^2}$ and $g'(x)>0$ for $x\in(0,1/2)$


and thus $g(x)\ge0$ for $x\in[0,1/2]$.
A: Here is a partial answer.
We have $\cos x \leq 1 - \frac{x^2}{2} + \frac{x^4}{24}$ for all $x$ because the cosine series is alternating.
Therefore, to suffices to prove that $\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)^{10} \le 1 -x^2$ for $x \in [0,0.5]$.
It is easy to verify that $f(x)=1 -x^2-\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)^{10} $ has a local minimum at $x=0$ and that $f(0)=0$.
The hard part is to prove that $f(x)\ge 0$ for $x \in [0,0.5]$. As we can see in the graph below, the local maximum is after $x=0.5$, but this seems harder to prove.

A: In order to prove
$$ 10\log\cos x\leq \log(1-x^2)\qquad \text{for }x\in\left[0,\tfrac{1}{2}\right]$$
it is enough to show
$$ 5 \tan x \geq \frac{x}{1-x^2} \qquad \text{for }x\in\left[0,\tfrac{1}{2}\right] $$
then use termwise integration. On the other hand $5(1-x^2)\tan(x)$ is a (log-)concave function on the given interval, hence its graph lies above the secant line through $(0,0)$ and $\left(\frac{1}{2}, \frac{15}{4}\tan\frac{1}{2}\right)$:
$$ 5(1-x)^2 \tan(x) \geq \frac{15}{2}\tan\left(\frac{1}{2}\right) x\geq\frac{15}{4}x\geq x $$
and the original inequality (which turns out to be pretty loose far from the origin) is proved.
A: Partial answer.
It equal to $(1-\sin^2x)^5\leq 1-x^2$.
$LHD≒1-5\sin^2x<1-x^2$, at $x≒0$.
Since $\sin x \leq x $, it must hold at $x=0.5$.
$LHD≒0.27$
$RHD=0.75$
