Question

If three positive real numbers $a,b,c$ are in A.P. such that $abc=4$, then the minimum value of $b$ is .

• A. $2^{\frac{1}{3}}$
• B. $2^{\frac{2}{3}}$
• C. $2^{\frac{1}{2}}$
• D. $2^{\frac{3}{2}}$

Answer 1

The correct option is B $2^{\frac{2}{3}}$

Since $a,b,c$ are in A.P., therfore, $b-a=d$ and $c-b=d$, where $d$ is the common difference of the A.P.

$\therefore a=b-d$ and $c=b+d$

Now,

$abc=4$

$\Rightarrow b(b-d)(b+d)=4$

$\Rightarrow \left[b\right(b^2-d^2\left)\right]=4$

But,

$b(b^2-d^2)\le b\times b^2$

$\Rightarrow b(b^2-d^2)\le b^3$

$\Rightarrow 4\le b^3$

$\Rightarrow b^3\ge 4$

$\Rightarrow b\ge 2^{\frac{2}{3}}$

Hence, the minimun value of b is $2^{\frac{2}{3}}$