Show that for all real numbers $a$ and $b$, $\,\, ab \le (1/2)(a^2+b^2)$ so as in the title, I have the following theorem to prove. 
Theorem
Show that for all $a$, $b\in \mathbb R$, that the following inequality holds,
$\begin{equation} ab \leq \frac{1}{2}(a^2 + b^2) \end{equation}$.
My attempt
So if $a=b=0$, then the inequality is trivially $0 \leq 0$ and we are ok.
If $a > 0$ and $b < 0$, then the product on the right is less than zero, while the one on the right is greater than $0$, so the inequality holds.
The problem that I am having is that if the numbers are the same sign, we have that both products will be positive and now we have to the show the inequality is true in this case. 
I tried making the following re-arrangement as a starting point $\begin{equation} \\ 2ab \leq (a^2 + b^2) \end{equation}$
But I am not really sure where to go from here. I've been staring at it for a while and I just can't seem to get anything out of it. I noticed that it looks rather similar to the law of cosines for theta equal to 90 degrees and $c = 0$, though that would be wierd indeed. Anyway that hasn't been talked about in the book yet so I don't think it matters. 
We have been introduced to the binomial theorem, difference of powers and geometric sum theorems, but I couldn't find any enlightenment from these.
I don't really want an answer if you can help it, but a hint in the right direction would be appreciated if you could offer one. Thanks.
 A: Hint: $\forall a,b \in \mathbb R, \; (a-b)^2\ge 0$
A: Remember that any real number squared is non-negative, and
$$a^2+b^2\geq 2ab\Longleftrightarrow a^2-2ab+b^2=(a-b)^2\geq 0\,\,...$$
A: Hint: Think about expanding and rearranging the inequality $(a-b)^2\geq 0$
A: Try using polar coordinates
$$a=r\cos \theta,b=r \sin \theta. $$
The inequality then becomes $$r^2 \cos \theta \sin \theta \leq \frac{1}{2}r^2, $$
or equivalently for nonzero $r$ $$2 \cos \theta \sin \theta (=\sin 2 \theta) \leq 1 .$$
A: Remember that for any $x \in \mathbb{R}$, we have that $x^2 \geq 0$ i.e. square of any real number is non-negative.
Hence, $$(a-b)^2 \geq 0$$ since $a,b \in \mathbb{R}$. Expand $(a-b)^2$ and rearrange to get what you want.
\begin{align}
(a-b)^2 & \geq 0\\
a^2 + b^2 - 2ab & \geq 0\\
a^2 + b^2 & \geq 2ab
\end{align}
A: We have a theorem in Classical Algebra that the arithmetic mean of n positive real numbers is greater than or equal to their geometric mean, the equality occurs when the numbers are equal.
In short, $A.M \geq G.M$
For any two real numbers $a, b,$ we know that $a^2,b^2$ are two positive real numbers. Now we apply the above stated theorem here.
$${{a^2+b^2}\over 2} \geq (a^2b^2)^{1\over2}\\ or,\space {{a^2+b^2}\over 2} \geq ab\\or, \space ab\leq{{a^2+b^2}\over 2}$$
This completes the proof.
