Maths Olympiad Prep

Track / Stage 7 / 31 of 300 #1431 of 1964

Problem 1431

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.0 Prove it

Example 3 Let a,b,ca, b, c be positive numbers, and satisfy abc=1a b c=1, prove:
2(a+1)2+b2+1+2(b+1)2+c2+1+2(c+1)2+a2+1⩽1\frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1} \leqslant 1

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

2(a+1)2+b2+1+2(b+1)2+c2+1+2(c+1)2+a2+1=2a2+b2+2a+2+2b2+c2+2b+2+2c2+a2+2c+2⩽22ab+2a+2+22bc+2b+2+22ca+2c+2=1ab+a+1+1bc+b+1+1ca+c+1=1,2(a+1)2+b2+1+2(b+1)2+c2+1+2(c+1)2+a2+1⩽1.\begin{array}{l} \frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1}= \frac{2}{a^{2}+b^{2}+2 a+2}+\frac{2}{b^{2}+c^{2}+2 b+2}+\frac{2}{c^{2}+a^{2}+2 c+2} \leqslant \\ \frac{2}{2 a b+2 a+2}+\frac{2}{2 b c+2 b+2}+\frac{2}{2 c a+2 c+2}= \\ \frac{1}{a b+a+1}+\frac{1}{b c+b+1}+\frac{1}{c a+c+1}=1, \\ \frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1} \leqslant 1 . \end{array}

Using the binary mean inequality x2+y2⩾2xyx^{2}+y^{2} \geqslant 2 x y, and problem 2, we can derive:
2(a+1)2+b2+1+2(b+1)2+c2+1+2(c+1)2+a2+1=2a2+b2+2a+2+2b2+c2+2b+2+2c2+a2+2c+2⩽22ab+2a+2+22bc+2b+2+22ca+2c+2=1ab+a+1+1bc+b+1+1ca+c+1=1,2(a+1)2+b2+1+2(b+1)2+c2+1+2(c+1)2+a2+1⩽1.\begin{array}{l} \frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1}= \frac{2}{a^{2}+b^{2}+2 a+2}+\frac{2}{b^{2}+c^{2}+2 b+2}+\frac{2}{c^{2}+a^{2}+2 c+2} \leqslant \\ \frac{2}{2 a b+2 a+2}+\frac{2}{2 b c+2 b+2}+\frac{2}{2 c a+2 c+2}= \\ \frac{1}{a b+a+1}+\frac{1}{b c+b+1}+\frac{1}{c a+c+1}=1, \\ \frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1} \leqslant 1 . \end{array}

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