Question

If a, b and c are positive real numbers, then the minimum value of
$\frac{2a^{2}b+2b^{2}c+2c^{2}a+2a{b}^2+2b{c}^2+2c{a}^2}{abc}$

• A. 6
• B. 8
• C. 12
• D. 36

Answer 1

The correct option is C 12

$\frac{2a^{2}b+2b^{2}c+2c^{2}a+2a{b}^2+2b{c}^2+2c{a}^2}{abc}$
$\frac{2a^{2}b+2a^{2}c+2b^{2}c+2b^{2}a+2c^{2}a+2c^{2}b}{abc}=\frac{2a^2(b+c)+2b^2(c+a)2c^2(a+b)}{abc}=\frac{2a(b+c)}{bc}+\frac{2a(a+c)}{ac}+\frac{2c(a+b)}{ab}=\frac{2a}{c}+\frac{2a}{b}+\frac{2a}{a}+\frac{2a}{c}+\frac{2a}{a}+\frac{2c}{b}=2(\frac{a}{c}+\frac{c}{a})+2(\frac{a}{b}+\frac{b}{a})+2(\frac{c}{b}+\frac{b}{c})$
$=12$