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Iranian Mathematical Olympiad

Algebra Difficulty 6.1 National Olympiad Prove it Iran

a, b and c are positive real numbers such that
∑cyc(a+b)2=2∑cyca+6abc. \sum_{cyc} (a+b)^2 = 2 \sum_{cyc} a + 6abc.
Prove that
∑cyc(a−b)2≤∣2∑cyca−6abc∣. \sum_{cyc} (a-b)^2 \le \left| 2 \sum_{cyc} a - 6abc \right|.

Solution

We know that ∑cyca2+∑cycab=∑cyca+3abc\sum_{cyc} a^2 + \sum_{cyc} ab = \sum_{cyc} a + 3abc, so
(∑cyca2+∑cycab)2=(∑cyca+3abc)2⇒(∑cyca2)2+(∑cycab)2+2(∑cyca2)(∑cycab)=(∑cyca)2+9a2b2c2+6abc∑cyca⇒(∑cyca2)2+(∑cycab)2−(∑cyca)2−9a2b2c2=6abc∑cyca−2(∑cyca2)(∑cycab). \begin{aligned} \left(\sum_{cyc} a^2 + \sum_{cyc} ab\right)^2 &= \left(\sum_{cyc} a + 3abc\right)^2 \\ \Rightarrow \quad &\left(\sum_{cyc} a^2\right)^2 + \left(\sum_{cyc} ab\right)^2 + 2\left(\sum_{cyc} a^2\right)\left(\sum_{cyc} ab\right) \\ &= \left(\sum_{cyc} a\right)^2 + 9a^2b^2c^2 + 6abc\sum_{cyc} a \\ \Rightarrow \quad &\left(\sum_{cyc} a^2\right)^2 + \left(\sum_{cyc} ab\right)^2 - \left(\sum_{cyc} a\right)^2 - 9a^2b^2c^2 \\ &= 6abc\sum_{cyc} a - 2\left(\sum_{cyc} a^2\right)\left(\sum_{cyc} ab\right). \end{aligned}
By AM-GM we have
(∑cyca2)(∑cycab)≥(∑cycab)2≥3abc∑cyca \left(\sum_{cyc} a^2\right) \left(\sum_{cyc} ab\right) \ge \left(\sum_{cyc} ab\right)^2 \ge 3abc \sum_{cyc} a
So
2(∑cyca2)(∑cycab)−6abc∑cyca≥6abc∑cyca−2(∑cyca2)(∑cycab) 2 \left(\sum_{cyc} a^2\right) \left(\sum_{cyc} ab\right) - 6abc \sum_{cyc} a \ge 6abc \sum_{cyc} a - 2 \left(\sum_{cyc} a^2\right) \left(\sum_{cyc} ab\right)

Then using above facts
2(∑cyca2)(∑cycab)−6abc∑cyca≥(∑cyca2)2+(∑cycab)2−(∑cyca)2−9a2b2c2⇔(∑cyca)2+9a2b2c2−6abc∑cyca≥(∑cyca2)2−2(∑cyca2)(∑cycab)+(∑cycab)2⇔(∑cyca−3abc)2≥(∑cyca2−∑cycab)2. \begin{aligned} & 2 \left( \sum_{cyc} a^2 \right) \left( \sum_{cyc} ab \right) - 6abc \sum_{cyc} a \\ & \qquad \geq \left( \sum_{cyc} a^2 \right)^2 + \left( \sum_{cyc} ab \right)^2 - \left( \sum_{cyc} a \right)^2 - 9a^2b^2c^2 \\ \Leftrightarrow & \left( \sum_{cyc} a \right)^2 + 9a^2b^2c^2 - 6abc \sum_{cyc} a \\ & \qquad \geq \left( \sum_{cyc} a^2 \right)^2 - 2 \left( \sum_{cyc} a^2 \right) \left( \sum_{cyc} ab \right) + \left( \sum_{cyc} ab \right)^2 \\ \Leftrightarrow & \left( \sum_{cyc} a - 3abc \right)^2 \geq \left( \sum_{cyc} a^2 - \sum_{cyc} ab \right)^2. \end{aligned}
Since ∑cyca2≥∑cycab\sum_{cyc} a^2 \geq \sum_{cyc} ab, we have ∑cyca2−∑cycab≥0\sum_{cyc} a^2 - \sum_{cyc} ab \geq 0. Therefore
∣∑cyca−3abc∣≥∑cyca2−∑cycab⇔∣2∑cyca−6abc∣≥∑cyc(a−b)2. \begin{aligned} \left| \sum_{cyc} a - 3abc \right| &\geq \sum_{cyc} a^2 - \sum_{cyc} ab \\ \Leftrightarrow &\left| 2 \sum_{cyc} a - 6abc \right| \geq \sum_{cyc} (a-b)^2. \end{aligned}

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