Question

If $\frac{(a+b+c)(ab+bc+ca)}{abc}=x$ where a, b, c are distinct real numbers, then x is always greater than

• A. 30
• B. 15
• C. 17
• D. 9

Answer 1

The correct option is D 9

Given: $\frac{(a+b+c)(ab+bc+ca)}{abc}=x$
We know that AM ≥ GM.
For any positive numbers a, b and c,
$\frac{a+b+c}{3}\ge (abc{)}^{\frac{1}{3}}$ ........(i)
and for any positive numbers ab, bc and ca,
$\frac{ab+bc+ca}{3}\ge (ab\times bc\times ca{)}^{\frac{1}{3}}$ ..........(ii)
$\Rightarrow \frac{ab+bc+ca}{3}\ge (abc{)}^{\frac{2}{3}}$
Multiplying (i) and (ii), we get,
$\frac{(a+b+c)(ab+bc+ca)}{9}\ge (abc{)}^{\frac{1}{3}}\times (abc{)}^{\frac{2}{3}}$
$\Rightarrow \frac{(a+b+c)(ab+bc+ca)}{9}\ge abc$
$\Rightarrow \frac{(a+b+c)(ab+bc+ca)}{9}\ge abc$
$\Rightarrow \frac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$
$\Rightarrow x\ge 9$
Hence, the correct answer is option(4).