We know that, A.M.> G.M.
∴ \(\frac{a\,+\,b}{2}\) > \(\sqrt{{a^2b^2}}\)
⇒ \(\frac{a^2\,+\,b^2}{2}\) > ab … (i)
Similarly \(\frac{b^2\,+\,c^2}{2}\) > \(\sqrt{{b^2c^2}}\)
⇒ \(\frac{b^2\,+\,c^2}{2}\) > bc ... (ii)
and \(\frac{c^2\,+\,a^2}{2}\) > \(\sqrt{{c^2a^2}}\)
⇒ \(\frac{c^2\,+\,a^2}{2}\) > ca …. (iii)
On adding eqs. (i), and (iii) we get
\(\frac{a^2\,+\,b^2}{2}\) + \(\frac{b^2\,+\,c^2}{2}\) + \(\frac{c^2\,+\,a^2}{2}\) > ab + bc + ca
⇒ a2 + b2 + c2 > ab + bc + ca
Hence proved