Q. For real numbers a,b,c if a+b+c=1 then prove that ab + bc + ca < or = 1/3.
Square both sides of $\displaystyle a+b+c=1$ and derive an expression for $\displaystyle a^{2}+b^{2}+c^{2}.$
Next, expand $\displaystyle (a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq 0,$ and derive an inequality for $\displaystyle a^{2}+b^{2}+c^{2}.$
Combine the two, and the required result follows.