Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$.
I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether $a$,$b$,$c$ and odd/even or is there a more general solution?

Another way:
$$a^2 + b^2 + c^2 = \frac{a^2 + b^2}{2} + \frac{b^2 + c^2}{2} + \frac{c^2 + a^2}{2} \ge ab + bc + ca$$
The last part follows from the fact that $x^2 + y^2 \ge 2xy$ for all $x, y$ (you can get this by observing $x^2 + y^2 - 2xy = (x - y)^2 \ge 0$ for all $x,y$).
