If $a x^2+\frac{b}{x} \geq c$ for all positive ' $x$ ' where $a0$ and $b0$ show that $27 a b^2 \ge

Question

If $a x^2+\frac{b}{x} \geq c$ for all positive ' $x$ ' where $a>0$ and $b>0$ show that $27 a b^2 \geq 4 c^3$.

Professor PC Thomas Classes · Answer

given $a x^2+\frac{b}{x} \geq c \forall x>0, \quad a>0, b>0$. to show that $27 a b^2 \geq 4 c^3$ Let us consider the function $f(x)=a x^2+\frac{b}{x}-c$ then $$ f^{\prime}(x)=2 a x-\frac{b}{x^2}=0 \Rightarrow x^3=\frac{b}{2 a} \Rightarrow x=\left(\frac{b}{2 a}ight)^{1 / 3} $$ Also $f^{\prime \prime}(x)=2 a+\frac{2 b}{x^3} \Rightarrow f^{\prime}\left(\left(\frac{b}{2 a}ight)^{1 / 3}ight)=2 a+\frac{2 b}{b} 	imes 2 a=6 a>0$ $	herefore \mathrm{f}$ is min at $x=\left(\frac{b}{2 a}ight)^{1 / 3}$ As (1) is true $\forall x \quad 	herefore$ is so for $x=\left(\frac{b}{2 a}ight)^{113}$ $$ \begin{aligned} & \Rightarrow a\left(\frac{b}{2 a}ight)^{2 / 3}+\frac{b}{(b / 2 a)^{1 / 3}} \geq c \ & \Rightarrow \frac{a\left(\frac{b}{2 a}ight)+b}{(b / 2 a)^{1 / 3}} \geq c \Rightarrow \frac{3 b}{2}\left(\frac{2 a}{b}ight)^{1 / 3} \geq c \end{aligned} $$ As a,b are + ve , cubing both sides, we get $$ \Rightarrow \frac{27 b^3}{8} \frac{2 a}{b} \geq c^3 \Rightarrow 27 a b^2 \geq 4 c^3 $$