Let $a, b, c$ are positive real numbers, such that $a+b+c=1$. Then prove that $\frac{a-bc}{a+bc} +\frac {b-ca}{b+ca} +\frac {c-ab}{c+ab} \leq \frac32$

Question

Let $a, b, c$ are positive real numbers, such that $a+b+c=1$. Then prove that $\frac{a-bc}{a+bc} + \frac{b-ca}{b+ca} + \frac{c-ab}{c+ab} \leq \frac32$

Who can help me ? I dont know what inequality need use in this case ??

Answer 1

Observations towards a solution. If you're stuck, explain what you've tried.

• If we normalize the denominator, we get $ a(a+b+c) + bc = a^2 + ab +ac + bc = (a+b)(a+c)$.
  • This allows us to easily combine denominators, so the simplification could be useful.
• If we normalize the numerator, then it's not as pretty. How can we get rid of the $-bc$?
  • We can get rid of it by adding 1 to the term.
  • WTS $\sum \frac{ 2a}{ (a+b)(a+c) } \leq \frac{ 9}{2}$
• Now let's make common denominator
  • WTS $\sum \frac{ 2a(b+c) } { (a+b)(a+c) } \leq \frac{ 9}{2} $.
• Now let's clear denominators and combine terms
  • WTS $8(ab+bc+ca) \leq 9 (a+b)(b+c)(c+a) $
• Now let's normalize the inequality
  • WTS $8(a+b+c) (ab+bc+ca) \leq 9(a+b)(b+c)(c+a)$.
• This is a well-known inequality.
  • Just expand and cancel terms, then apply AM-GM.
  • Hence equality holds iff $ a = b = c = \frac{1}{3}$ .

Answer 2

We have $$\begin{align} LHS &=\sum_{cyc}\frac{a-bc}{a+bc}\\ &= \sum_{cyc}\frac{a-bc}{1-(b+c)+bc}\\ &= \sum_{cyc}\frac{a-bc}{(1-b)(1-c)}\\ &= \frac{\sum_{cyc}\left((a-bc)(1-a)\right)}{(1-b)(1-c)(1-c)}\\ &= \frac{\sum_{cyc}\left(a-a^2-bc+abc\right)}{(1-b)(1-c)(1-c)}\\ &= \frac{1-\sum_{cyc}a^2 -\sum_{cyc}ab+3abc}{1-\sum_{cyc}a +\sum_{cyc}ab-abc}\\ &=\frac{\sum_{cyc}ab+3abc}{\sum_{cyc}ab-abc} \end{align}$$ then $$\begin{align} LHS \le \frac{3}{2}&\iff 2\left(\sum_{cyc}ab+3abc\right) \le 3\left(\sum_{cyc}ab-abc \right)\\ &\iff \sum_{cyc}ab \ge 9abc\\ &\iff \sum_{cyc}a \cdot \sum_{cyc}ab \ge 9abc \\ &\iff \sum_{cyc}a^2b + \sum_{cyc}ab^2 + 3abc \ge 9abc \\ &\iff \sum_{cyc}a^2b + \sum_{cyc}ab^2 \ge 6abc \tag{1} \end{align}$$ Applying the AM-GM inequality, we can prove that $(1)$ holds true: $$\sum_{cyc}a^2b \ge 3\sqrt[3]{a^2b\cdot b^2c\cdot c^2a} = 3abc $$ $$\sum_{cyc}ab^2 \ge 3\sqrt[3]{ab^2\cdot bc^2\cdot ca^2} = 3abc $$

The equality occurs if and only if $a=b=c= 1/3$.

Q.E.D