# 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.

Note by Rishabh Deep Singh
2 years, 4 months ago

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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!

- 2 years, 4 months ago