Question

Given 3 positive reals $a,b,c$, find the minimum value of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}$.

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{7}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{2}$

we can observe that for $a=b=c=1$
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}$ is minimum and $=\frac{3}{2}$
Therefore answer is $A$