Question

If $a,b$ and $c>0$, then the minimum value of $\frac{a}{(b+c)}+\frac{b}{(c+a)}+\frac{c}{(a+b)}$ is

• A. $1$
• B. $\frac{3}{2}$
• C. $2$
• D. $\frac{5}{2}$

Answer 1

The correct option is B

$\frac{3}{2}$

Explanation for the correct option:

Step 1. Find the minimum value of $\frac{a}{(b+c)}+\frac{b}{(c+a)}+\frac{c}{(a+b)}$:

As we know, $AM\ge GM$

$\frac{\left[{\displaystyle \frac{a}{(b+c)}}+{\displaystyle \frac{b}{(c+a)}}+{\displaystyle \frac{c}{(a+b)}}\right]}{3}\ge \frac{∛abc}{(a+b)(b+c)(c+a)}$ — (1)

Also, $\frac{[a+b]}{2}\ge \surd ab,\frac{[b+c]}{2}\ge \surd bc$ and $\frac{[c+a]}{2}\ge \surd ca$

$\Rightarrow$$(a+b)(b+c)(c+a)\ge 8abc$

$\Rightarrow$$\frac{abc}{(a+b)(b+c)(c+a)}\le \frac{1}{8}$

Step 2. Put this value in equation(1):

$\frac{a}{(b+c)}+\frac{b}{(c+a)}+\frac{c}{(a+b)}\ge 3\sqrt[3]{\frac{1}{8}}$

$\mathbf{\therefore }\frac{\mathbf{a}\mathbf{}}{\mathbf{(}\mathbf{b}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{)}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{b}\mathbf{}}{\mathbf{(}\mathbf{c}\mathbf{}\mathbf{+}\mathbf{}\mathbf{a}\mathbf{)}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{c}}{\mathbf{(}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}\mathbf{)}}\mathbf{}\mathbf{\ge }\mathbf{}\frac{\mathbf{3}}{\mathbf{2}}$

Hence, Option ‘B’ is Correct.