If a, b, c are distinct positive real numbers and a^2+b^2+c^2=1, then ab+bc+ca is

Question

If $a, b, c$ are distinct positive real numbers and $a^2+b^2+c^2=1$, then $ab+bc+ca$ is

Accepted Answer

As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$

$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$

$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$

$$ \Rightarrow ab + bc + ca < 1$$

Solutions · 1