Question

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

• A. $ab<\frac{c^2}{4}$
• B. $ab\ge \frac{c^2}{4}$
• C. $ab\ge \frac{c}{4}$
• D. None of these

Answer 1

The correct option is B $ab\ge \frac{c^2}{4}$

Let $f\left(x\right)=ax+\frac{b}{x}-c$; $x>0;a,b>0$
$\Rightarrow$ $f^{\prime }\left(x\right)=a-\frac{b}{x^2}$
$f^{\prime }\left(x\right)=0$
$\Rightarrow$ $a-\frac{b}{x^2}=0$
$\Rightarrow$ $x={\left(\frac{b}{a}\right)}^{\frac{1}{2}}$

But $ax+\frac{b}{x}\ge c$.
$\therefore$ $f\left(x\right)\ge 0$ for all $x>0$
$\Rightarrow f\left[\right({\frac{b}{a})}^{\frac{1}{2}}]\ge 0$
$\Rightarrow a{\left(\frac{b}{a}\right)}^{\frac{1}{2}}+b{\left(\frac{a}{b}\right)}^{\frac{1}{2}}-c\ge 0$
$\Rightarrow 2\sqrt{ab}\ge c$
$\Rightarrow ab\ge \frac{c^2}{4}$