Must Install - Surah Yaseen Android App

# Conversation Between Islami Bhai and jahid89

### 19 Visitor Messages

1. MORAL of the Story : We should have friends everywhere for not only sharing joyful moments but for support in rainy days also .
2. i willtry to solve ur question..=)
3. 4. (a+b)(b+c)(c+a) =0
= (ab + ac + bb + bc) (c+a)
= abc + acc + bbc + bcc + aab + aac + abb + abc
= 2abc + a(bb + cc) + b(aa + cc) + c(aa + bb)
= 2abc + 6abc - 6abc + a(bb + cc) + b(aa + cc) + c(aa + bbc)
= 8abc + a(bb - 2bc + cc) + b(aa - 2ac + cc) + c(aa - 2ab + bb)
= 8abc + [a(b-c)^2 + b(a-c)^2 + c(a-b)^2]

therefore (a+b)(b+c)(c+a) = 8abc + X
where X = [a(b-c)^2 + b(a-c)^2 + c(a-b)^2] >= 0

since 8abc + X >= 8abc,
(a+b)(b+c)(c+a) >= 8abc
5. ok no problem...
i will try to solve
6. dont think so i can solve the question u posted 7. i havent reached the right answer so thats y was asking for it
8. one qus for you..
if (a+b)(b+c)(c+a)=o then prove that
(a+b)(b+c)(c+a) >= 8abc
9. hummmmmmmm ok u r right..
so what is the solution...
10. but where did the q go while finding the constant k
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