Prove $a^2 + b^2 \geq ab$ and $a^2 + b^2 \geq -ab$ I know that $a^2 + b^2 \geq 2ab$, and I have been trying to use that to prove $a^2  + b^2 \geq ab$, but I haven't made any progress on using that. Can I have a hint?


To prove $a^{2} + b^{2} \geq ab$ for any real numbers $a$ and $b$, regardless of if they are positive or negative:

Note that when you square a number, you get a nonnegative number, so this implies $(a - b)^{2} \geq 0$.  
But $(a - b)^{2} = a^{2} -2ab + b^{2}$, so we get $a^{2} + b^{2} \geq 2ab$.  But:
$$a^{2} + b^{2} \geq 2ab \implies \frac{1}{2}(a^{2} + b^{2}) \geq ab $$
But $a^{2} + b^{2} \geq \frac{1}{2}(a^{2} + b^{2})$ since half of a positive number is smaller than that positive number (and $a^{2} + b^{2}$ is positive), so this gives $$a^{2} + b^{2}\geq ab. $$


Also, to show $a^{2} + b^{2} \geq -ab$:

Since a number squared is always nonnegative, $(a+b)^{2} \geq 0$.
But $(a +b)^{2} = a^{2} + 2ab + b^{2}$, so $a^{2} + 2ab + b^{2} \geq 0$, which implies $a^{2} + b^{2} \geq -2ab$.  Multiplying both sides by $\frac{1}{2}$ gives:  $$\frac{1}{2}(a^{2} + b^{2}) \geq -ab $$ and again, as above, since $a^{2} + b^{2} \geq \frac{1}{2}(a^{2} + b^{2})$, we get the desired result: $$ a^{2} + b^{2} \geq -ab.$$
