Application Of Derivatives Ques 113
113. If $a x^{2}+b / x \geq c$ for all positive $x$ where $a>0$ and $b>0$, then show that $27 a b^{2} \geq 4 c^{3}$.
$(1982,2 \mathrm{M})$
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Solution:
Formula:
- Given, $a x^{2}+\frac{b}{x} \geq c, \forall x>0 ; a, b>0$
Let
$ f(x)=a x^{2}+\frac{b}{x}-c $
$\therefore \quad f^{\prime}(x)=2 a x-\frac{b}{x^{2}}=\frac{2 a x^{3}-b}{x^{2}}$
$\Rightarrow f^{\prime \prime}(x)=2 a+\frac{2 b}{x^{3}}>0$ [since, $a, b$ are all positive]
Now, put $\quad f^{\prime}(x)=0 \Rightarrow x=(\frac{b}{2 a})^{1 / 3}>0 \quad[\because a, b>0]$
At $\quad x=(\frac{b}{2 a})^{1 / 3}, f^{\prime \prime}(x)=+\mathrm{ve}$
$\Rightarrow f(x)$ has minimum at $x=(\frac{b}{2 a})^{1 / 3}$.
$ \begin{array}{rlrl} \text { and } \quad f \quad ((\frac{b}{2 a}))^{1 / 3}= & a (\frac{b}{2 a})^{2 / 3}+\frac{b}{(b / 2 a)^{1 / 3}}-c \geq 0 \\ & =(\frac{2 a}{b}){ }^{1 / 3} \cdot \frac{3 b}{2}-c \geq 0 \\ \Rightarrow & & (\frac{2 a}{b}){ }^{1 / 3} \cdot \frac{3 b}{2} \geq c \end{array} $
On cubing both sides, we get
$ \begin{array}{rlrl} & & \frac{2 a}{b} \cdot \frac{27 b^{3}}{8} \geq c^{3} \\ \Rightarrow & 27 a b^{2} \geq 4 c^{3} \end{array} $
BETA
