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Verify by Mean Value Theory or otherwise that $\cos{x}<\{\frac{\sin{x}}{x}\}^3$ for $0

I am unable to solve the problem. Please give me a solution of the problem.

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    Do you mean something specific with $\{\}$ or are they just normal brackets? – 2017-01-08
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    @Arthur Just normal brackets – 2017-01-08
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    But $\lim_{x\to 0^+ } \cos x =(\frac{\sin x }{x})^3 $ which is 1=1 :( – 2017-01-08
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    @Ramanujan limits do not play any role here. – 2017-01-08

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We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt[3]{\cos{x}}}-x$.

Indeed, by AM-GM $$f'(x)=\frac{\cos{x}\sqrt[3]{\cos{x}}+\frac{\sin^2x}{3\sqrt[3]{\cos^2x}}}{\sqrt[3]{\cos^2x}}-1=\frac{3\cos^2x+\sin^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=$$ $$=\frac{1+\cos^2x+\cos^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}\geq\frac{3\sqrt[3]{\cos^4x}-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=0.$$ Thus, $f(x)>f(0)=0$ and we are done!

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    why is $1+2\cos^2 x\geq 3\sqrt[3]{\cos^4 x}$ for $x\in(0,\frac{\pi}{2})$? – 2017-01-08
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    @Max It's AM-GM or we can do the following thing. Let $\sqrt[3]{\cos^2x}=t$. Hence, we need to prove that $2t^3+1-3t^2\geq0$, which is $(t-1)^2(2t+1)\geq0$. – 2017-01-08
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    @MichaelRozenberg Please elaborate AM-GM that you have applied – 2017-01-08
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    @user1942348 I did do it already: $1+\cos^2x+\cos^2x\geq3\sqrt[3]{\cos^4x} $ by AM-GM. – 2017-01-08
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For $x > 0$ the well-known representation as alternating series give the estimates $$ \sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} > x - \frac{x^3}{6} \, ,\\ \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} < 1 - \frac{x^2}{2} + \frac{x^4}{24} \, . $$

For $0 < x < \pi/2$ in particular $x^2 < 6$ holds and therefore $$ \left( \frac{\sin x}{x} \right )^3 - \cos x > \left( 1 - \frac{x^2}{6} \right )^3 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} \right ) \\ = \frac{x^4}{24}- \frac{x^6}{216} = \frac{x^4}{24} \left( 1 - \frac{x^2}{9} \right) > 0 \, . $$