Question

If $a+b+c=1$, the greatest value of $\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}$, is

• A. $\frac{1}{3}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{2}$

Given that, $a+b+c=1$

We know that $A.M.\ge H.M.$

$⟹\frac{a+b}{2}\ge \frac{2ab}{a+b}$

$⟹\frac{ab}{a+b}\le \frac{a+b}{4}$ ————-(1)

Similarly, $\frac{bc}{b+c}\le \frac{b+c}{4}$ ————-(2)

and $\frac{ac}{a+c}\le \frac{a+c}{4}$ ————-(3)

adding (1), (2) and (3) we get

$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\le \frac{a+b+b+c+a+c}{4}$

$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\le \frac{2(a+b+c)}{4}$

$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\le \frac{1}{2}$

Therefore, the greatest value of $\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{bc}{b+c}$ is $\frac{1}{2}$