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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?

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

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

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  • $\begingroup$ You can simplify the criterion for equality: $a + b = a + c = b + c \iff a = b = c$. $\endgroup$ – Clement Yung May 20 '20 at 1:05
  • $\begingroup$ Sure, I intended to leave as it is. $\endgroup$ – Zhanxiong May 20 '20 at 1:15
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Well, by Cauchy-Schwarz, $$((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge (1+1+1)^2=9$$ $$\iff (a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge \frac{9}{2} $$

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