If ax + frac{b}{x} ge c for all positive x,wh | vedclass

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(B) Let $f(x) = ax + \frac{b}{x} - c$,where $x > 0$ and $a, b > 0$.To find the minimum value,we differentiate $f(x)$ with respect to $x$:$f'(x) = a -

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

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(B) Let $f(x) = ax + \frac{b}{x} - c$,where $x > 0$ and $a, b > 0$.To find the minimum value,we differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{b}{x^2} = \frac{ax^2 - b}{x^2}$.Setting $f'(x) = 0$,we get $ax^2 = b$,which implies $x = \sqrt{\frac{b}{a}}$ (since $x > 0$).The function $f(x)$ attains its minimum value at this point.Substituting $x = \sqrt{\frac{b}{a}}$ into $f(x)$:$f\left(\sqrt{\frac{b}{a}}\right) = a\sqrt{\frac{b}{a}} + \frac{b}{\sqrt{b/a}} - c = \sqrt{ab} + \sqrt{ab} - c = 2\sqrt{ab} - c$.According to the problem,$f(x) \ge 0$ for all $x > 0$,so the minimum value must be $\ge 0$.Therefore,$2\sqrt{ab} - c \ge 0$,which means $2\sqrt{ab} \ge c$.Squaring both sides,we get $4ab \ge c^2$,or $ab \ge \frac{c^2}{4}$.

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

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