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If $a \in (\pi/2, \pi)$, then $\sqrt{x^2+x}+\frac{\tan^2a}{\sqrt{x^2+x}}$ is always greater than or equal to

(A) $2\tan a$
(B) $1$
(C) $-1$
(D) $-2\tan a$

I know that the $\tan(\cdot)$ function takes its minimum value as $0$. Tried to use this. But I'm not able to eliminate $x$ from my answer.

Please help

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Hint: For positive numbers $a$ and $b$, $$a+b \ge 2\sqrt{ab}.$$

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Hint:   use the AM-GM inequality, then consider the sign of $\tan a\,$:

$$\require{cancel} \dfrac{\sqrt{x^2+x} + \dfrac{\tan^2 a}{\sqrt{x^2+x}}}{2} \ge \sqrt{\cancel{\sqrt{x^2+x}} \cdot \dfrac{\tan^2 a}{\cancel{\sqrt{x^2+x}}}} = |\tan a| $$

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  • $\begingroup$ Thank you so much. It was my stupidity that AM GM inequality didn't strike my mind..... $\endgroup$ – Shraman Maiti Jul 10 '18 at 6:26

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