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Algebra Difficulty 5.0 AIME, harder Prove it Romania

Consider v,wv, w two distinct non-zero complex numbers. Prove that
zw+wˉzv+vˉ, |zw + \bar{w}| \leq |zv + \bar{v}|,
for any zCz \in \mathbb{C}, z=1|z| = 1, if and only if there exists k[1,1]k \in [-1, 1] such that w=kvw = kv.

Solution

If there exists k[1,1]k \in [-1, 1] such that w=kvw = kv, the inequality is obvious. Conversely, let t>1t > 1 such that wtv0w - tv \ne 0. Setting
z=tvˉwˉwtv, z = \frac{t\bar{v} - \bar{w}}{w - tv},
we have z=1|z| = 1 and moreover
zw+wˉ=t(wvˉwˉv)wtv, zw + \bar{w} = \frac{t(w\bar{v} - \bar{w}v)}{w - tv},
zv+vˉ=wvˉwˉvwtv, zv + \bar{v} = \frac{w\bar{v} - \bar{w}v}{w - tv},
implying
zw+wˉ=tzv+vˉ. |zw + \bar{w}| = t |zv + \bar{v}|.
As t>1t > 1, the given inequality yields zw+wˉ=zv+vˉ=0|zw + \bar{w}| = |zv + \bar{v}| = 0, so wvˉwˉv=0w\bar{v} - \bar{w}v = 0, therefore wv=kR\frac{w}{v} = k \in \mathbb{R}.
Plugging back in the initial condition we derive that k[1,1]k \in [-1, 1], as needed.

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