Let $a,$ $b,$ and $t$ be real numbers such that $a + b = t.$  Find, in terms of $t,$ the minimum value of $a^2 + b^2.$
We can rewrite $a^2 + b^2$ as $(a+b)^2 - 2ab$. And we are given that $a+b = t$. So, $a^2 + b^2 = t^2 - 2ab$. So, the problem is now to minimize $t^2 - 2ab$. But the only restriction we have is that $a + b = t$. Let's try to find the maximum of $ab$ by using the AM-GM inequality. The AM-GM inequality states that the arithmetic mean of a set of nonnegative numbers is always greater or equal to the geometric mean. For our case, it says that $\frac{a+b}{2} \geq \sqrt{ab}$. But we are given that $a+b = t$. So the inequality becomes $\frac{t}{2} \geq \sqrt{ab}$. If we square both sides we get $t^2 \geq 4ab$. So, $ab \leq \frac{t^2}{4}$. So, the maximum value of $ab$ is $\frac{t^2}{4}$. So, the minimum value of $t^2 - 2ab$ is $t^2 - 2\frac{t^2}{4}$. This simplifies to $\frac{t^2}{2}$. So $\frac{t^2}{2}$ is our answer.

