JEE MAIN - Mathematics (2002 - No. 43)
If $$a,\,b,\,c$$ are distinct $$ + ve$$ real numbers and $${a^2} + {b^2} + {c^2} = 1$$ then $$ab + bc + ca$$ is
less than 1
equal to 1
greater than 1
any real no.
Paliwanag
As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$
$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$
$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$
$$ \Rightarrow ab + bc + ca < 1$$
$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$
$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$
$$ \Rightarrow ab + bc + ca < 1$$