Question

if a/b+c=b/c+a=c/a+b than prove that (a^2/b+c)+(b^2/c+a)+(c^2/a+b)=0

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Answer 1

Correct Question :--- if a/(b+c) + b/(c+a) + c/(a+b) = 1 ,, than prove that (a^2/b+c)+(b^2/c+a)+(c^2/a+b) = 0 ?

Solution :---

→ a/(b+c) + b/(c+a) + c/(a+b) = 1

Multiply LHS and RHS by (a+b+c) , we get,

→ a(a+b+c)/(b+c) + b(a+b+c)/(c+a) + c(a+b+c)/(a+b) = (a+b+c)

Now, lets see Numerator part First,

→ a(a+b+c) can be written as a²+a(b+c)= [ a²/(b+c) + a ]

similarly,

→ b(a+b+c) = b² + b(c+a)

→ c(a+b+c) = c² + c(a+b)

Putting these values in numerator , we can write ,

→ a² + a(b+c)/(b+c) = [ a²/(b+c) + a ]

→ b² + b(c+a)/(c+a) = [ b²/(c+a) + b ]

→ c² + c(a+b)/(a+b) = [ c²/(a+b) + c ]

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Putting these values now , we get,

→ [ a²/(b+c) + a ] + [ b²/(c+a) + b ] + [ c²/(a+b) + c ] = (a+b+c)

or,

→ [ a²/(b+c) + b²/(c+a) + c²/(a+b) ] + (a+b+c) = (a+b+c)

Now, we can see that, (a+b+c) will be cancel from Each side,

So, we get,

→ [ a²/(b+c) + b²/(c+a) + c²/(a+b) ] = 0

✪✪ Hence Proved ✪✪