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Question: If α∈\(\left( 0,\frac{\pi}{2} \right)\)then \(\sqrt{x^{2} + x} + \frac{\tan^{2}\alpha}{\sqrt{x^{2} +...

If α∈(0,π2)\left( 0,\frac{\pi}{2} \right)then x2+x+tan⁡2αx2+x\sqrt{x^{2} + x} + \frac{\tan^{2}\alpha}{\sqrt{x^{2} + x}}is always greater than or equal to

A

2 tan σ

B

1

C

1

D

sec2 α

Answer

2 tan σ

Explanation

Solution

Let a=x2+x 6muand 6mub=tan⁡2αx2+xa = \sqrt{x^{2} + x}\mspace{6mu} and\mspace{6mu} b = \frac{\tan^{2}\alpha}{\sqrt{x^{2} + x}}

then using AM ≥ GM, we get, a+b2≥ab\frac{a + b}{2} \geq \sqrt{ab}

⇒ a + b ≥ 2ab\sqrt{ab}

⇒x2+x+tan⁡2αx2+x≥2tan⁡2α=2 6mutan⁡ 6muα\sqrt{x^{2} + x} + \frac{\tan^{2}\alpha}{\sqrt{x^{2} + x}} \geq 2\sqrt{\tan^{2}\alpha} = 2\mspace{6mu}\tan\mspace{6mu}\alpha

[... α ∈ (0, π/2)]