Question 55
a, b and c are positive real numbers such that
$$
\sum_{cyc} (a+b)^2 = 2 \sum_{cyc} a + 6abc.
$$
Prove that
$$
\sum_{cyc} (a-b)^2 \le \left| 2 \sum_{cyc} a - 6abc \right|.
$$
$$
\sum_{cyc} (a+b)^2 = 2 \sum_{cyc} a + 6abc.
$$
Prove that
$$
\sum_{cyc} (a-b)^2 \le \left| 2 \sum_{cyc} a - 6abc \right|.
$$