Question

If $ax+\frac{b}{x}\le$ 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. $ab=c/4$

Answer 1

The correct option is B

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

Let $f\left(x\right)=ax+\frac{b}{x}-c;x>0;a,b>0f^{\prime }\left(x\right)=0\Rightarrow a-\frac{b}{x^2}=0\Rightarrow x={\left(\frac{b}{a}\right)}^{1/2}Butax+\frac{b}{x}\le c;\therefore f\left(x\right)\ge 0forallx>0\therefore f\left[{\left(\frac{b}{a}\right)}^{1/2}\right]\ge 0\Rightarrow a{\left(\frac{b}{a}\right)}^{1/2}+b{\left(\frac{a}{b}\right)}^{1/2}-c\ge 0\Rightarrow 2\sqrt{ab}\ge c\Rightarrow ab\ge \frac{c^2}{4}.$
Hence (b) is the correct answer.