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Problem 1666

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.7 Prove it

Deepening 3 Let $a, b, c$ be positive real numbers, $-2<\lambda<2$ , prove:
$\begin{array}{l} \sqrt{\left(a^{2}+\lambda a b+b^{2}\right)\left(b^{2}+\lambda b c+c^{2}\right)}+ \\ \sqrt{\left(b^{2}+\lambda b c+c^{2}\right)\left(c^{2}+\lambda c a+a^{2}\right)}+ \\ \sqrt{\left(c^{2}+\lambda c a+a^{2}\right)\left(a^{2}+\lambda a b+b^{2}\right)} \geqslant\left(a^{2}+b^{2}+\right. \\ \left.c^{2}\right)+(1-\lambda)(a b+b c+c a) . \end{array}$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Prove that by implementing formula transformation, we get
$a^{2}+\lambda a b+b^{2}=\frac{2-\lambda}{4}(a+b)^{2}+\frac{2+\lambda}{4}(a-b)^{2},$

Thus, we can construct the complex numbers
$\begin{array}{l} z_{1}=\frac{\sqrt{2-\lambda}}{2}(a+b)+\frac{\sqrt{2+\lambda}}{2}(a-b) i, \\ z_{2}=\frac{\sqrt{2-\lambda}}{2}(b+c)+\frac{\sqrt{2+\lambda}}{2}(b-c) i, \\ z_{3}=\frac{\sqrt{2-\lambda}}{2}(c+a)+\frac{\sqrt{2+\lambda}}{2}(c-a) i . \end{array}$

It is easy to calculate that $z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=\left(a^{2}+b^{2}+c^{2}\right)+$
$(1-\lambda)(a b+b c+c a) .$

Therefore, the left side of inequality (9) $=\left|z_{1}\right|\left|z_{2}\right|+$
$\begin{array}{l} \left|z_{2}\right|\left|z_{3}\right|+\left|z_{3}\right|\left|z_{1}\right| \\ \quad=\left|z_{1} z_{2}\right|+\left|z_{2} z_{3}\right|+\left|z_{3} z_{1}\right| \\ \quad \geqslant\left|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right| \\ \quad=\left|\left(a^{2}+b^{2}+c^{2}\right)+(1-\lambda)(a b+b c+c a)\right| \\ \quad \geqslant\left(a^{2}+b^{2}+c^{2}\right)+(1-\lambda)(a b+b c+c a) \end{array}$

Hence, inequality (9) is proved.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.