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Cartesian Product

ntIf, ab/1+bc...

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ntIf, ab/1+bc+ bc/1+ca +ca/1+ab=1n nt Then prove that 1/a+1/b+1/c= 62n

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a b + b c + c a = 0

then prove that

\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b} = 0

Q. If 1/a, 1/b, 1/c, are in a.p prove that BC, ca, an are in a.p
hints: prove that ab- ca =ca- BC
Using 1/c - 1/b = 1/b - 1/a.

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{1}{\sqrt{b c}} + \frac{1}{\sqrt{c a}} + \frac{1}{\sqrt{a b}}

A + B + C = 180

then prove that

cot B cot C + cot C cot A + cot A cot B

\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in A.P., prove that
(i) bc,ca,ab are in A.P.
(ii) a(b+c), b(c+a), c(a+b) are in A.P

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