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Distributivity

137. a+b+c=0 ...

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137. a+b+c=0 , (a(b-c)) +(b(c-a))+(c(a-b))=0 prove that

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\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}

a (b - c) + b (c - a) + c (a - b) = 0

Q. It is known that |a| = |b| = |c| = 1 and a + b + c = 0. Prove that ab + bc + ca = -3/2.

Q. If a^2+b^2+c^2-ab-bc-ca =0,prove that a=b=c.

a^{2} + b^{2} + c^{2} - a b - b c - c a = 0

a = b = c

\frac{b c}{a} + \frac{c a}{b} + \frac{a b}{c} \geq a + b + c,

a, b, c > 0

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