If $\alpha\epsilon\left( 0,\frac{\pi}{2} ight)$, then $\sqrt{x^{2} + x} + \frac{ an^{2}\alpha}{\sqrt{x^{2} + x}}$ is always greater than or equal to

Question

If $\alpha\epsilon\left( 0,\frac{\pi}{2} ight)$, then $\sqrt{x^{2} + x} + \frac{	an^{2}\alpha}{\sqrt{x^{2} + x}}$ is always greater than or equal to

Dr. Abubacker · Accepted Answer

(a)
Here,$\alpha \in \left( 0,\frac{\pi}{2} ight) \Rightarrow 	an{\alpha\ is\ ( + ve)}$
[as, we know if $a,b > 0 \Rightarrow \frac{a + b}{2} \geq \sqrt{ab}\ ie,\ AM \geq GM$]
$\frac{\sqrt{x^{2} + x +}\frac{	an^{2}\alpha}{\sqrt{x^{2} + x}}}{2} \geq \sqrt{\sqrt{x^{2} + x.}\frac{	an^{2}\alpha}{\sqrt{x^{2} + x}}\lbrack using\ AM\  \geq GMbrack}$
$\Rightarrow \sqrt{x^{2} + x} + \frac{	an^{2}\alpha}{\sqrt{x^{2} + x}} \geq 2	an\alpha$

Q. If $\alpha\epsilon\left( 0,\frac{\pi}{2} \right)$, then $\sqrt{x^{2} + x} + \frac{\tan^{2}\alpha}{\sqrt{x^{2} + x}}$ is always greater than or equal to

Explanation

Reviewed by Dr. Abubacker

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