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Inequality from China

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Source: lemondian(https://kuing.cjhb.site/thread-13667-1-1.html)

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sqing

#1 h Apr 3, 2025, 1:11 PM • 5 Y

Y by cubres, aidan0626, Quantum-Phantom, rayford, SunnyEvan

Let $x\in (0,\frac{\pi}{2}) .$ Prove that $tanx\ge x^k$ Where $k=1,2,3,4.$

This post has been edited 1 time. Last edited by sqing, Apr 3, 2025, 1:12 PM

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#2 h Apr 3, 2025, 3:28 PM

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sqing wrote:

Let $x\in (0,\frac{\pi}{2}) .$ Prove that $tanx\ge x^k$ Where $k=1,2,3,4.$

When k=1:
Let $f(x)=tan(x)-x$
$f'(x)= sec^2(x)-1=tan^2(x)>0\iff f(x)>0$ When k=2:
Let $f(x)=tan(x)-x^2$
$f'(x)= sec^2(x)-2x>x^2-2x+1\geq 0\iff f(x)>0$

This post has been edited 2 times. Last edited by SunnyEvan, Apr 3, 2025, 3:40 PM

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Quantum-Phantom

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Quantum-Phantom

#3 h Apr 4, 2025, 2:21 AM

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Proof of $\tan x\ge x^4$ : It suffices to prove that $f(x)=\arctan x^4-x\le0$ for $x\in\left(0,\tfrac\pi2\right)$ . We have $f'(x)=\tfrac{4x^3-x^8-1}{x^8+1}$ . Let $g(x)=4x^3-x^8-1$ , then $g'(x)=4x^2\left(3-2x^5\right)$ , which means that $g(x)$ increases on $\left(0,\sqrt[5]{\tfrac32}\right)$ and decreases on $\left(\sqrt[5]{\tfrac32},+\infty\right)$ . Thus, $f'(x)=0\Leftrightarrow g(x)=0$ has two positive roots $x_1<\sqrt[5]{\frac32}<x_2$ . By the intermediate value theorem, we get $\tfrac35<x_1<\tfrac23$ and $\tfrac97<x_2<\tfrac43$ . It suffices to prove that $f(0)$ , $f(x_1)$ , $f(x_2)\le0$ . Note that $f(0)=0$ and
$f(x_1)=\arctan x_1^4-x_1\le\arctan\frac{16}{81}-\frac35<\frac\pi6-\frac35<0,$ where we used $\tfrac{16}{81}<\tan\tfrac\pi6=\tfrac1{\sqrt3}$ and $\pi<3$ ; as well as
$f(x_2)=\arctan x_2^4-x_2\le\arctan\frac{256}{81}-\frac97<0,$ where the last inequality sign is equivalent to $\tan\tfrac97>\tfrac{256}{81}$ . Indeed,
$\tan\frac97-\frac{256}{81}=\int_0^1\frac{x(1-x)\left(107595 x+3911-14283 x^2\right)\sin\frac{9x}7}{83349\cos\frac97}\operatorname dx > 0.$

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sqing

#4 h Apr 4, 2025, 3:16 AM • 1 Y

Y by SunnyEvan

Thank you.

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