# Let $a$, $b$, and $c$ be positive real numbers. What is the smallest possible value of…

Let $$a$$, $$b$$, and $$c$$ be positive real numbers. What is the smallest possible value of $$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)$$?

I don't know how to approach this problem, though I think it might use the AM-GM inequality. Can someone please help?

• Hint: The inequality is homogeneous so we may assume $a+b+c=1.$ – Display name May 20 '20 at 0:44

Hint: Do you know why $$( a + b+ c) ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) \geq 9$$?

Arithmetic Mean - Harmonic Mean.

Hence, conclude that $$2 ( a + b + c) ( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} ) \geq 9$$

When does equality occur?

• Where does the factor 2 comes from. – hamam_Abdallah May 20 '20 at 0:52
• Because you require the same numbers as there are in the denominator, $(a+b)+(b+c)+(c+a)= 2(a+b+c)$. – MathAnimal May 20 '20 at 1:02
• Alternatively, let $x = a+b, y = b+c, z = c+a$ to see the transformation. – Calvin Lin May 20 '20 at 1:04

AM-GM inequality is a good idea: \begin{align} & (a + b + c)\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\ = & \frac{1}{2}((a + b) + (a + c) + (b + c))\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\ \geq & \frac{1}{2} \times 3\sqrt[3]{(a + b)(a + c)(b + c)} \times 3\sqrt[3]{\frac{1}{(a + b)(a + c)(b + c)}} \\ = & \frac{9}{2} \end{align}

The equality holds when $$a + b = a + c = b + c$$.

• You can simplify the criterion for equality: $a + b = a + c = b + c \iff a = b = c$. – Clement Yung May 20 '20 at 1:05
• Sure, I intended to leave as it is. – Zhanxiong May 20 '20 at 1:15

Well, by Cauchy-Schwarz, $$((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge (1+1+1)^2=9$$ $$\iff (a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge \frac{9}{2}$$