Question:
If three positive real numbers a, b, c are in A.P. and abc = 4, then the minimum possible value of b is_________.
Solution:
Let us suppose three positive real numbers a, b and c are in A.P and abc = 4
Since arithmetic mean ≥ geometric mean
i. e $\frac{a+b+c}{3} \geq \sqrt[3]{a b c}$
i. e $\frac{a+b+c}{3} \geq \sqrt[3]{4}$ (given)
Since $b=\frac{a+c}{2}$
$2 b=a+c$
i. e $\frac{2 b+b}{3} \geq(4)^{\frac{1}{3}}$
i.e $\frac{3 b}{3} \geq(4)^{\frac{1}{3}}$
i. e $b \geq(2)^{\frac{2}{3}}$
i.e minimum possible value of $b$ is $(2)^{\frac{2}{3}}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.