Question

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

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

(b) $ab\frac{>}{}\frac{c^2}{4}$

(c) $ab\frac{>}{}\frac{c^{}}{4}$

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)ab\ge \frac{c^2}{4} \\ \mathrm{Given}:ax+\frac{b}{x}\ge c \\ \mathrm{Minimum}\mathrm{value}\mathrm{of}ax+\frac{b}{x}=c \\ \mathrm{Now}, \\ f\left(x\right)=ax+\frac{b}{x} \\ \Rightarrow f\text{'}\left(x\right)=a-\frac{b}{x^2} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow a-\frac{b}{x^2}=0 \\ \Rightarrow a{x}^2-b=0 \\ \Rightarrow a{x}^2=b \\ \Rightarrow x^2=\frac{b}{a} \\ \Rightarrow x=\pm \frac{\sqrt{b}}{\sqrt{a}} \\ f\text{'}\text{'}\left(x\right)=\frac{2b}{x^3} \\ \Rightarrow f\text{'}\text{'}\left(x\right)=\frac{2b}{{\left({\displaystyle \frac{\sqrt{b}}{\sqrt{a}}}\right)}^3} \\ \Rightarrow f\text{'}\text{'}\left(x\right)=\frac{2b{\left(a\right)}^{{\displaystyle \frac{3}{2}}}}{{\left(b\right)}^{{\displaystyle \frac{3}{2}}}}>0 \\ \mathrm{So},x=\frac{\sqrt{b}}{\sqrt{a}}\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{minima}. \\ \therefore f\left(\frac{\sqrt{b}}{\sqrt{a}}\right)=a\left(\frac{\sqrt{b}}{\sqrt{a}}\right)+\frac{b}{\left({\displaystyle \frac{\sqrt{b}}{\sqrt{a}}}\right)}\ge c \\ =\sqrt{a}\sqrt{a}\left(\frac{\sqrt{b}}{\sqrt{a}}\right)+\frac{\sqrt{b}\sqrt{b}}{\left({\displaystyle \frac{\sqrt{b}}{\sqrt{a}}}\right)}\ge c \\ =\sqrt{ab}+\sqrt{ab}\ge c \\ \Rightarrow 2\sqrt{ab}\ge c \\ \Rightarrow \frac{c}{2}\le \sqrt{ab} \\ \Rightarrow \frac{c^2}{4}\le ab \end{gathered}$$