Olympiad Maths Prep

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Problem 1624

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.5 Prove it

Example 24 (Self-created problem, 2006. 12.17) Let a,b,c,d∈R−a, b, c, d \in \mathbf{R}^{-}, and ac=1a c=1, then
aa+ab+1+bb+bc+1+cc+cd+1+dd+da+1⩽43\frac{a}{a+a b+1}+\frac{b}{b+b c+1}+\frac{c}{c+c d+1}+\frac{d}{d+d a+1} \leqslant \frac{4}{3}

Equality in (38) holds if and only if a=b=c=d=1a=b=c=d=1.

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

aa+ab+1+bb+bc+1+cc+ca+1=1−(1−abc)2(a+ab+1)(b+bc+1)(c+ca+1)\begin{array}{l} \frac{a}{a+a b+1}+\frac{b}{b+b c+1}+\frac{c}{c+c a+1}= \\ 1-\frac{(1-a b c)^{2}}{(a+a b+1)(b+b c+1)(c+c a+1)} \end{array}

This is because
 The left side of the above equation =aa+ab+1+aba+ab+1+1a+ab+1+bb+bc+1−aba+ab+1+cc+ca+1−1a+ab+1=1+b(1−abc)(b+bc+1)(a+ab+1)−1−abc(c+ca+1)(a+ab+1)=1−1−abca+ab+1(1c+ca+1−bb+bc+1)=1−(1−abc)2(a+ab+1)(b+bc+1)(c+ca+1)\begin{aligned} \text { The left side of the above equation }= & \frac{a}{a+a b+1}+\frac{a b}{a+a b+1}+\frac{1}{a+a b+1}+\frac{b}{b+b c+1}- \\ & \frac{a b}{a+a b+1}+\frac{c}{c+c a+1}-\frac{1}{a+a b+1}= \\ & 1+\frac{b(1-a b c)}{(b+b c+1)(a+a b+1)}-\frac{1-a b c}{(c+c a+1)(a+a b+1)}= \\ & 1-\frac{1-a b c}{a+a b+1}\left(\frac{1}{c+c a+1}-\frac{b}{b+b c+1}\right)= \\ & 1-\frac{(1-a b c)^{2}}{(a+a b+1)(b+b c+1)(c+c a+1)} \end{aligned}

Thus,
 The left side of equation (38) =(aa+ab+1+bb+bc+1+cc+ca+1)+(cc+cd+1+dd+da+1+aa+ac+1)−(cc+ac+1+aa+ac+1)=2−(1−abc)2(a+ab+1)(b+bc+1)(c+ca+1)−(1−cda)2(c+cd+1)(d+da+1)(a+ac+1)−[1−1+ac+(ac)2(a+ac+1)(c+ac+1)]=1−(1−b)2(a+ab+1)(b+bc+1)(2+c)−(1−d)2(c+cd+1)(d+da+1)(2+a)+3(2+a)(2+c)⩽1+3(2+a)(2+c)⩽1+13=43\begin{aligned} \text { The left side of equation (38) }= & \left(\frac{a}{a+a b+1}+\frac{b}{b+b c+1}+\frac{c}{c+c a+1}\right)+ \\ & \left(\frac{c}{c+c d+1}+\frac{d}{d+d a+1}+\frac{a}{a+a c+1}\right)- \\ & \left(\frac{c}{c+a c+1}+\frac{a}{a+a c+1}\right)= \\ & 2-\frac{(1-a b c)^{2}}{(a+a b+1)(b+b c+1)(c+c a+1)}- \\ & \frac{(1-c d a)^{2}}{(c+c d+1)(d+d a+1)(a+a c+1)}- \\ & {\left[1-\frac{1+a c+(a c)^{2}}{(a+a c+1)(c+a c+1)}\right]=} \\ & 1-\frac{(1-b)^{2}}{(a+a b+1)(b+b c+1)(2+c)}- \\ & \frac{(1-d)^{2}}{(c+c d+1)(d+d a+1)(2+a)}+ \\ & \frac{3}{(2+a)(2+c)} \leqslant 1+\frac{3}{(2+a)(2+c)} \leqslant \\ & 1+\frac{1}{3}=\frac{4}{3} \end{aligned}

Therefore, equation (38) holds, and equality is achieved if and only if a=b=c=d=1a=b=c=d=1.

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