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Algebra Difficulty 5.0 AIME Prove it Ukraine

Prove that for any positive real numbers aa, bb, c<1c < 1 satisfying
2(a+b+c)+4abc=3(ab+bc+ca)+1, 2(a+b+c) + 4abc = 3(ab+bc+ca) + 1,
the following inequality holds: a+b+c≤34a+b+c \le \frac{3}{4}.

Solution

We can rewrite the given equality as follows:
4abc−4(ab+bc+ca)+4(a+b+c)−4=−(ab+bc+ca)+2(a+b+c)−3,4abc - 4(ab + bc + ca) + 4(a+b+c) - 4 = -(ab + bc + ca) + 2(a+b+c) - 3,
4(a−1)(b−1)(c−1)=−(a−1)(b−1)−(b−1)(c−1)−(c−1)(a−1),4(a-1)(b-1)(c-1) = -(a-1)(b-1) - (b-1)(c-1) - (c-1)(a-1),
4=11−a+11−b+11−c.4 = \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}.
The function f(x)=11−xf(x) = \frac{1}{1-x} is convex on the interval x<1x < 1 because f′′(x)=2(1−x)3≥0f''(x) = \frac{2}{(1-x)^3} \ge 0. So, by the Jensen's inequality,
4=f(a)+f(b)+f(c)≥3f(a+b+c3)=93−(a+b+c). 4 = f(a) + f(b) + f(c) \ge 3f\left(\frac{a+b+c}{3}\right) = \frac{9}{3-(a+b+c)}.
The last inequality implies the required one.

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