Help for Integration (3)

$\large m \sin x + \int_0^x (\sec t)^m \, dt > (m+1) x$

For all $0 <x<\dfrac\pi2$ , and $m \in \mathbb N$ , the inequality above holds true.

Prove the inequality above without differentiating it.

#Calculus

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Cool question!
Only an outline, sorry, because I'm stretched for time.

Write $(\sec t)^m$ as $(1 + \tan^2 t)^{\frac{m}{2}}$ .

Then use Bernoulli's inequality, $(1 + x)^n > 1 + nx$ for $x = \tan^2 t, n = \frac{m}{2}$ .

Integrate the terms (not difficult), and then you'll get, this is equivalent to, $2 \sin x + \tan x > 3x$ which is true (if you see the Maclaurin series expansions).

EDIT: Screw Maclaurin series, I found this absolute beauty of a gem to prove $2 \sin x + \tan x > 3x$ .

$2 \sin x + \tan x = \int\limits_{0}^{x} (2\cos t + \sec^2 t) dt = \int\limits_{0}^{x} \left(\cos t + \cos t + \dfrac{1}{\cos^2 t}\right) dt > \int\limits_{0}^{x} 3 dt = 3x$
where the inequality follows by AM-GM!

Ameya Daigavane - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$