Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

81. (Original problem, 2007.05.13) Let a,b,c∈Ra, b, c \in \mathbf{R}, and ∑a2=1\sum a^{2}=1, then
∑1−bc⩾7−∑bc\sum \sqrt{1-b c} \geqslant \sqrt{7-\sum b c}

Equality holds if and only if a=b=ca=b=c.

Solution

81. Proof: Since
21−ab⋅1−ac=2∑a2−ab⋅∑a2−ac=∑a2+c2+(a−b)2⋅∑a2+b2+(a−c)2⩾∑a2+bc+(a−b)(a−c)=1+bc+(a−b)(a−c)\begin{aligned} 2 \sqrt{1-a b} \cdot \sqrt{1-a c}= & 2 \sqrt{\sum a^{2}-a b} \cdot \sqrt{\sum a^{2}-a c}= \\ & \sqrt{\sum a^{2}+c^{2}+(a-b)^{2}} \cdot \sqrt{\sum a^{2}+b^{2}+(a-c)^{2}} \geqslant \\ & \sum a^{2}+b c+(a-b)(a-c)= \\ & 1+b c+(a-b)(a-c) \end{aligned}

Similarly, there are two other inequalities. Therefore,
(1−bc+1−ca+1−ab)2=3−∑bc+2∑1−ca1−ab⩾3−∑bc+3+∑bc+∑(a−b)(a−c)=7−∑bc\begin{array}{l} (\sqrt{1-b c}+\sqrt{1-c a}+\sqrt{1-a b})^{2}= \\ 3-\sum b c+2 \sum \sqrt{1-c a} \sqrt{1-a b} \geqslant \\ 3-\sum b c+3+\sum b c+\sum(a-b)(a-c)= \\ 7-\sum b c \end{array}

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