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?
 A: 
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: $$ a^{2} + b^{2} \geq -ab.$$
A: Hint: $a^2\pm ab+b^2=\frac{1}{2}(a^2+b^2+(a\pm b)^2)$
A: For a different approach, note that both $$\left(a+\frac b2\right)^2+\frac {3b^2}4\ge 0$$ and $$\left(a-\frac b2\right)^2+\frac {3b^2}4\ge 0$$
(obtained by completing the square)
A: If $ab\ge 0$, then $a^2+b^2\ge 2ab\ge ab$.
If $ab\le 0$, then $a^2+b^2\ge0\ge ab$.

As this holds for all signs of $a$ and $b$, it also holds for $-ab$.
A: Look at $\;a^2\pm ab+b^2\;$ as a quadratic in $\;a\;$ , so its discriminant is
$$\Delta=b^2-4b^2=-3b^2\le0$$
If $\;b=0\;$ both inequalities are trivial, otherwise the above shows both quadratics (one for each sign) have no real roots and thus mantain all the time the same sign, and since for $\;a=b\;$ we get $\;b^2>0\;\;\text{or}\;\;3b^2>0\;$ , the expressions are always positive.
A: If you can show that $ab \geq 0$, then $2ab \geq ab$, and you can use transitivity to conclude the proof.
