Question

If $a+b+c=1$, where $a,b,c$ are positive real numbers, then the minimum value of $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$ is

• A. $13$
• B. $24$
• C. $27$
• D. $32$

Answer 1

The correct option is C $27$

$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=\frac{1}{abc}$
We know that,
$A.M.\ge G.M\Rightarrow \frac{a+b+c}{3}\ge (abc{)}^{1/3}\Rightarrow {\left(\frac{1}{3}\right)}^3\ge abc\Rightarrow \frac{1}{abc}\ge 27$

Therefore,
$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\ge 27$

Hence, the minimum value of the given expression is $27$.