Minimum of the expression

Algebra Level 3

For positive real numbers a , b , c a,b,c the minimum of a b + c + b c + a + c a + b \dfrac a{b+c}+\dfrac b{c+a}+\dfrac c{a+b} is


The answer is 1.5.

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1 solution

Put b + c = x b+c=x , c + a = y c+a=y , a + b = z a+b=z . Then a = 1 2 ( x + y + z ) a=\dfrac12(-x + y + z) , b = 1 2 ( x y + z ) b=\dfrac12(x - y + z) , c = 1 2 ( x + y z ) c=\dfrac12(x + y - z) . So a b + c + b c + a + c a + b = 1 2 x ( x + y + z ) + 1 2 y ( x y + z ) + 1 2 z ( x + y z ) = 1 2 ( 3 + y x + x y + z x + x z + y z + z y ) 1 2 ( 3 + 2 + 2 + 2 ) = 3 2 \dfrac a{b+c}+\dfrac b{c+a}+\dfrac c{a+b}=\dfrac1{2x}(-x + y + z)+\dfrac1{2y}(x - y + z)+\dfrac1{2z}(x + y - z) =\dfrac12\Big(-3+\dfrac yx+\dfrac xy+\dfrac zx+\dfrac xz+\dfrac yz+\dfrac zy\Big)\geq\dfrac12(-3+2+2+2)=\dfrac32 . Since x , y , z > 0 x,y,z>0 , the minimal value of the expressions y x + x y , z x + x z , y z + z y \dfrac yx+\dfrac xy, \dfrac zx+\dfrac xz, \dfrac yz+\dfrac zy is 2 2 , and the minimal value of the given expression is 3 2 \dfrac32 .

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