If a, b and c be the positive numbers, then prove that a2+b2+c2 is gre

1.	If a, b and c be the positive numbers, then prove that a2+b2+c2 is greater than ab + bc + ca.
Answer» 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

If a, b and c be the positive numbers, then prove that a2+b2+c2 is greater than ab + bc + ca.

Answer»

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