Track / Stage 6 / 128 of 400 #1128 of 2000
Problem 1128 National olympiad, first round Difficulty 6.2 Prove it
Example 2 Let a , b , c a, b, c a , b , c a b ( a + c ) ( b + c ) + b c ( b + c ) ( c + a ) \frac{a b}{(a+c)(b+c)}+\frac{b c}{(b+c)(c+a)} ( a + c ) ( b + c ) ab + ( b + c ) ( c + a ) b c + c a ( c + b ) ( a + b ) ⩾ 3 4 +\frac{c a}{(c+b)(a+b)} \geqslant \frac{3}{4} + ( c + b ) ( a + b ) c a ⩾ 4 3
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Official solution 4 [ a b ( a + b ) + b c ( b + c ) + c a ( c + a ) ] ⩾ 3 ( a + b ) ( b + c ) ( c + a ) ⇔ 4 [ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) ] ⩾ 3 [ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) + 2 a b c ] ⇔ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) ⩾ 6 a b c . \begin{array}{l}
4[a b(a+b)+b c(b+c)+c a(c+a)] \geqslant 3(a+b)(b+c)(c+a) \\
\Leftrightarrow 4\left[a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)\right] \\
\geqslant 3\left[a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)+2 a b c\right] \\
\Leftrightarrow a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right) \geqslant 6 a b c .
\end{array} 4 [ ab ( a + b ) + b c ( b + c ) + c a ( c + a )] ⩾ 3 ( a + b ) ( b + c ) ( c + a ) ⇔ 4 [ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) ] ⩾ 3 [ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) + 2 ab c ] ⇔ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) ⩾ 6 ab c .
By the AM-GM inequality, we have (1) ⩾ a ( 2 b c ) + b ( 2 c a ) + c ( 2 a b ) = 6 a b c \geqslant a(2 b c)+b(2 c a)+c(2 a b)=6 a b c ⩾ a ( 2 b c ) + b ( 2 c a ) + c ( 2 ab ) = 6 ab c
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Source: NuminaMath-1.5 ,
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