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Show that $$\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x,\forall x\in \left ( 0;\frac\pi2 \right )$$


Firstly, I had use the differentiation of $f(x)=\left (\frac{\sin x}{x} \right )^3- \cos^{2}x$,but I think it very complex: $$f'(x)=3\left ( \frac{\sin x}{x} \right )^2\frac{x\cos x-\sin x}{x^2}+\sin2x$$ Or I can't prove it be positive in $ \left ( 0;\frac\pi2 \right )$

Secondly, I had use the convex function, but function $f(x)=\left (\frac{\sin x}{x} \right )^3-\cos^{2}x$ isn't convex or concave in all $ \left ( 0;\frac\pi2 \right )$

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    $\begingroup$ what's your attempt? $\endgroup$
    – haqnatural
    Aug 30 '15 at 13:28
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    $\begingroup$ Please wait, I will add it $\endgroup$
    – mja
    Aug 30 '15 at 13:29
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It is enough to consider: $$ f(x)= \sin^3(x)-x^3\cos^2(x) $$ and prove it is an increasing function over $I=\left(0,\frac{\pi}{2}\right)$ by studying: $$ g(x)=\frac{f'(x)}{\cos(x)} = -3x^2\cos(x)+2x^3\sin(x)+3\sin^2(x).$$ $g(x)> 0$ follows from $\frac{\sin x}{x}> 1-\frac{x^2}{6}$ and $\cos(x)< 1$ over $I$.

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    $\begingroup$ Not obvious why $g(x) \ge 0$ $\endgroup$
    – user261263
    Aug 30 '15 at 13:55
  • $\begingroup$ @EugenCovaci: with the last inequalities the problem boils down to checking a polynomial inequality, namely: $$x^2-\frac{x^4}{4}\geq 0.$$ $\endgroup$ Aug 30 '15 at 13:57
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Since all the quantities involved are positive, you can shift around the terms to get an equivalent inequality: $$\tan^2x\sin x\ge x^3\tag{i}$$ Now consider the function $f(x)=\tan^2x\sin x - x^3$.

Its derivative is $f'(x)=\tan^2 x\cos x+2\tan x \sin x\sec^2 x-3x^2=g(x)-3x^2$.

Using the AM-GM inequality on the $g(x)$, we get : $$g(x)\ge 3\tan^2x(\sec x)^{1/3}\tag{ii}$$

From (ii) we can say that $$f'(x)\ge 3(\tan^2x(\sec x)^{1/3}-x^2)\tag{iii}$$

Finally using the inequalities $\tan x>x$ and $\sec x >1$ on (iii) we get that $f'(x)\ge0$.

Now that it is known that f(x) is an increasing function, $$f(x)>f(0)\implies \tan^2x\sin x\ge x^3$$

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Start with $$ 1-\cos(x)\ge0\tag{1} $$ For $x\ge0$, successive integration yields $$ x-\sin(x)\ge0\tag{2} $$ and $$ \frac{x^2}2-1+\cos(x)\ge0\tag{3} $$ and $$ \frac{x^3}6-x+\sin(x)\ge0\tag{4} $$ and $$ \frac{x^4}{24}-\frac{x^2}2+1-\cos(x)\ge0\tag{5} $$ From $(4)$, we get $$ \frac{\sin^3(x)}{x^3}\ge1-\frac{x^2}2+\frac{x^4}{12}-\frac{x^6}{216}\tag{6} $$ From $(5)$, we get $$ \cos(x)\le1-\frac{x^2}2+\frac{x^4}{24}\tag{7} $$ Subtracting $(7)$ from $(6)$ yields $$ \frac{\sin^3(x)}{x^3}-\cos(x)\ge\frac{x^4}{24}-\frac{x^6}{216}\tag{8} $$ and for $|x|\le3$, $\frac{x^4}{24}-\frac{x^6}{216}\ge0$. Thus for, $x\in\left[0,\frac\pi2\right]$, $$ \frac{\sin^3(x)}{x^3}\ge\cos(x)\ge\cos^2(x)\tag{9} $$

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We can have a stronger one which is easier:

Let $f(x)=\tan{x}\sin{x} -x^2,x\in[0,\pi/2),$

$$f'(x)>0\implies \tan{x}\sin{x}>x^2,x\in(0,\pi/2).$$

$$f'(x)=\sin{x}\dfrac{\cos^2{x}+1}{\cos^2{x}}-2x>0\iff \dfrac{\tan{x}}{x}> \dfrac{2}{\cos{x}+\dfrac{1}{\cos{x}}} $$

which is true, because:

$$\dfrac{\tan{x}}{x}>1,\cos{x}+\dfrac{1}{\cos{x}}>2 \implies \dfrac{2}{\cos{x}+\dfrac{1}{\cos{x}}}<1,\forall x\in(0,\pi/2).$$

it is trivial that

$$\dfrac{\tan{x}}{x} \cdot\tan{x}\sin{x}>x^2,x\in(0,\pi/2).$$

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