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Algebra Difficulty 6.1 National olympiad Prove it

Problem 1 (1963 Moscow Mathematical Olympiad Question) Given that a,b,ca, b, c are positive numbers, prove: ab+c+bc+a+ca+b⩾32\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}.

Solution

Proof: Without loss of generality, assume a⩾b⩾ca \geqslant b \geqslant c,
∵ab+c+bc+a+ca+b−32=(ab+c−12)+(bc+a−12)+(ca+b−12)=2a−b−c2(b+c)+2b−a−c2(c+a)+2c−a−b2(a+b)⩾2a−b−c2(c+a)+2b−a−c2(c+a)+2c−a−b2(a+b)=a+b−2c2(a+c)+2c−a−b2(a+b)⩾a+b−2c2(a+b)+2c−a−b2(a+b)=a+b−2c+2c−a−b2(a+b)=0.\begin{aligned} \because & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2} \\ = & \left(\frac{a}{b+c}-\frac{1}{2}\right)+\left(\frac{b}{c+a}-\frac{1}{2}\right)+\left(\frac{c}{a+b}-\frac{1}{2}\right) \\ = & \frac{2 a-b-c}{2(b+c)}+\frac{2 b-a-c}{2(c+a)}+\frac{2 c-a-b}{2(a+b)} \\ \geqslant & \frac{2 a-b-c}{2(c+a)}+\frac{2 b-a-c}{2(c+a)}+\frac{2 c-a-b}{2(a+b)} \\ = & \frac{a+b-2 c}{2(a+c)}+\frac{2 c-a-b}{2(a+b)} \\ \geqslant & \frac{a+b-2 c}{2(a+b)}+\frac{2 c-a-b}{2(a+b)} \\ = & \frac{a+b-2 c+2 c-a-b}{2(a+b)} \\ = & 0 . \end{aligned}
∴\therefore The original inequality holds.

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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.