# If the solve the problem

Question:

If $a x+\frac{b}{x} \geq c$ for all positive $x$ where $a, b,>0$, then

(a) $a b<\frac{c^{2}}{4}$

(b) $a b \geq \frac{c^{2}}{4}$

(c) $a b \geq \frac{c}{4}$

Solution:

$(\mathrm{b}) a b \geq \frac{c^{2}}{4}$

Given : $a x+\frac{b}{x} \geq c$

Minimum value of $a x+\frac{b}{x}=c$

Now,

$f(x)=a x+\frac{b}{x}$

$\Rightarrow f^{\prime}(x)=a-\frac{b}{x^{2}}$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow a-\frac{b}{x^{2}}=0$

$\Rightarrow a x^{2}-b=0$

$\Rightarrow a x^{2}=b$

$\Rightarrow x^{2}=\frac{b}{a}$

$\Rightarrow x=\pm \frac{\sqrt{b}}{\sqrt{a}}$

$f^{\prime \prime}(x)=\frac{2 b}{x^{3}}$

$\Rightarrow f^{\prime \prime}(x)=\frac{2 b}{\left(\frac{\sqrt{b}}{\sqrt{a}}\right)^{3}}$

$\Rightarrow f^{\prime \prime}(x)=\frac{2 b(a)^{\frac{3}{2}}}{(b) \frac{3}{2}}>0$

So, $x=\frac{\sqrt{b}}{\sqrt{a}}$ is a local minima.

$\therefore f\left(\frac{\sqrt{b}}{\sqrt{a}}\right)=a\left(\frac{\sqrt{b}}{\sqrt{a}}\right)+\frac{b}{\left(\frac{\sqrt{b}}{\sqrt{a}}\right)} \geq c$

$=\sqrt{a} \sqrt{a}\left(\frac{\sqrt{b}}{\sqrt{a}}\right)+\frac{\sqrt{b} \sqrt{b}}{\left(\frac{\sqrt{b}}{\sqrt{a}}\right)} \geq c$

$=\sqrt{a b}+\sqrt{a b} \geq c$

$\Rightarrow 2 \sqrt{a b} \geq c$

$\Rightarrow \frac{c}{2} \leq \sqrt{a b}$

$\Rightarrow \frac{c^{2}}{4} \leq a b$