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

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

Prove the inequality above without differentiating it.

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

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

Prove the inequality above without differentiating it.

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TopNewestCool 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 · 6 months ago

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