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# if a , b , c are three positive real numbers then prove that a/(b+c) + b/(c+a) + c/(a+c) >= 3/2 .

Akshay
185 Points
6 years ago
(a/(b+c)) = [(a+b+c)/(b+c)] – 1;      eq(1)
//ly (b/(c+a)) = [(a+b+c)/(a+c)] – 1;    eq(2)
//ly (c/(a+b)) = [(a+b+c)/(a+b)] – 1;    eq(3)

now, { [(a+b+c)/(b+c)] + [(a+b+c)/(a+c)] + [(a+b+c)/(a+b)] } / 3 >= 3 / { (b+c)/(a+b+c) + (a+c)/(a+b+c) + (a+b)/(a+b+c) }                          AM>=HM

Multiply by 3 on both sides,

{ [(a+b+c)/(b+c)] + [(a+b+c)/(a+c)] + [(a+b+c)/(a+b)] }  >= 9 /{ (b+c)/(a+b+c) + (a+c)/(a+b+c) + (a+b)/(a+b+c) }

{ (b+c)/(a+b+c) + (a+c)/(a+b+c) + (a+b)/(a+b+c) } = 2 , you can solve for that

so,
{ [(a+b+c)/(b+c)] + [(a+b+c)/(a+c)] + [(a+b+c)/(a+b)] }  >= 9/2

subtact 3 and use eq(1) , eq(2) and eq(3). and u get r result