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# If a X + B X > C for All Positive X Where A,B,>0, Then - Mathematics

#### Question

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

##### Options
• ab<c^2/4

• ab>=c^2/4

• ab>=c/4

#### Solution

$\ ab \geq \frac{c^2}{4}$

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

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

$\text { Now },$

$f\left( x \right) = ax + \frac{b}{x}$

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

$\text { For a local maxima or a local minima, we must have}$

$f'\left( x \right) = 0$

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

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

$\Rightarrow a x^2 = b$

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

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

$f''\left( x \right) = \frac{2b}{x^3}$

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

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

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

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

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

$= \sqrt{ab} + \sqrt{ab} \geq c$

$\Rightarrow 2\sqrt{ab} \geq c$

$\Rightarrow \frac{c}{2} \leq \sqrt{ab}$

$\Rightarrow \frac{c^2}{4} \leq ab$

Is there an error in this question or solution?