Question

prove that a2+b2+c2÷ab+bc+ca greater than 1

Answer 1

to prove
${a}^{2} + {b}^{2} + {c}^{2} > ab + bc + ca$
but
${a}^{2} + {b}^{2} + {c}^{2} = {(a + b + c)}^{2} - 2(ab + bc + ca) > ab + bc + ca$
there fore cutting ab+bc+ca from both sides.
(a+b+c)^2-2>1

hence proved a square + b square + c square + divided by a b + BC + CA is greater than 1 .

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