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

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 $x\in[0,1/2]$.
