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Difficulty 7.2 National olympiad, round 2 Prove it
Example 2 Let a , b , c a, b, c a , b , c a b c = 1 a b c=1 ab c = 1 2 ( a + 1 ) 2 + b 2 + 1 + 2 ( b + 1 ) 2 + c 2 + 1 + 2 ( c + 1 ) 2 + a 2 + 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} \leqslant 1
\end{array} ( a + 1 ) 2 + b 2 + 1 2 + ( b + 1 ) 2 + c 2 + 1 2 + ( c + 1 ) 2 + a 2 + 1 2 ⩽ 1
Solution 2 ( a + 1 ) 2 + b 2 + 1 + 2 ( b + 1 ) 2 + c 2 + 1 + 2 ( c + 1 ) 2 + a 2 + 1 = 2 a 2 + b 2 + 2 a + 2 + 2 b 2 + c 2 + 2 b + 2 + 2 c 2 + a 2 + 2 c + 2 ⩽ 2 2 a b + 2 a + 2 + 2 2 b c + 2 b + 2 + 2 2 c a + 2 c + 2 = 1 a b + a + 1 + 1 b c + b + 1 + 1 c a + c + 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 .
\end{array} ( a + 1 ) 2 + b 2 + 1 2 + ( b + 1 ) 2 + c 2 + 1 2 + ( c + 1 ) 2 + a 2 + 1 2 = a 2 + b 2 + 2 a + 2 2 + b 2 + c 2 + 2 b + 2 2 + c 2 + a 2 + 2 c + 2 2 ⩽ 2 ab + 2 a + 2 2 + 2 b c + 2 b + 2 2 + 2 c a + 2 c + 2 2 = ab + a + 1 1 + b c + b + 1 1 + c a + c + 1 1 = 1.
Prove that using the binary mean inequality x 2 + y 2 ⩾ 2 x y x^{2}+y^{2} \geqslant 2 x y x 2 + y 2 ⩾ 2 x y 2 ( a + 1 ) 2 + b 2 + 1 + 2 ( b + 1 ) 2 + c 2 + 1 + 2 ( c + 1 ) 2 + a 2 + 1 = 2 a 2 + b 2 + 2 a + 2 + 2 b 2 + c 2 + 2 b + 2 + 2 c 2 + a 2 + 2 c + 2 ⩽ 2 2 a b + 2 a + 2 + 2 2 b c + 2 b + 2 + 2 2 c a + 2 c + 2 = 1 a b + a + 1 + 1 b c + b + 1 + 1 c a + c + 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 .
\end{array} ( a + 1 ) 2 + b 2 + 1 2 + ( b + 1 ) 2 + c 2 + 1 2 + ( c + 1 ) 2 + a 2 + 1 2 = a 2 + b 2 + 2 a + 2 2 + b 2 + c 2 + 2 b + 2 2 + c 2 + a 2 + 2 c + 2 2 ⩽ 2 ab + 2 a + 2 2 + 2 b c + 2 b + 2 2 + 2 c a + 2 c + 2 2 = ab + a + 1 1 + b c + b + 1 1 + c a + c + 1 1 = 1.
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