Here is what I had to prove.
Question: For positive reals $a$ and $b$ prove that $a^2+b^2 \geq 2ab$.
Here is how my teacher did it:
First assume that it is in fact, true that $a^2+b^2 \geq 2ab$. Therefore $a^2+b^2-2ab \geq 0$ . We have $(a-b)^2$ is greater than or equal to zero which is true. Hence what was assumed originally is true.
Why does this method work?
I cannot understand how you can first assume that it is true and then prove that it is.
As written your teacher's proof is incorrect. The correct way to write your teacher's proof would be to write $$a^2+b^2 \geq 2ab\iff a^2+b^2-2ab \geq 0\iff(a-b)^2\geq 0$$ where each of the statements are equivalent to each other. The final statement is true, hence so is the first. It is much cleaner to write the proof in the reverse direction i.e. $$(a-b)^2\geq 0\implies a^2+b^2-2ab \geq 0\implies a^2+b^2 \geq 2ab$$
Your teacher is incorrect. In order to prove something is true, you do not assume it first. Here is a correct, logical way.
Let $a$ and $b$ be real numbers. Then, $(a-b)^{2}\geq 0$. Hence, $a^{2}-2ab+b^{2}\geq 0$, and thus, $a^{2}+b^{2}\geq 2ab$.
If the teacher worded it exactly as you have, then he is absolutely wrong.
However, there's a proof technique that follows a similar structure and is correct, which might have been what the teacher was going for. I'm talking about having the inference direction opposite to the writing direction.
Usually a proof starts with something that is true, makes a deduction from it, and continues until you reach your result: $a \Rightarrow b \Rightarrow c \Rightarrow d$.
Instead, you can start with what you want to prove, but instead of showing what follows from it, show what it follows from: $d \Leftarrow c \Leftarrow b \Leftarrow a$. If you end up with something true, then your original claim is true as well.
This is what your teacher was trying to do: $a^2+b^2\ge 2ab \Leftarrow a^2+b^2-2ab \ge 0 \Leftarrow (a-b)^2 \ge 0$. The last statement is true, so this completes the proof.
Some of the confusion might arise from the fact that in this case, all these inferences are reversible - it's true that $a^2+b^2\ge 2ab \Leftarrow a^2+b^2-2ab \ge 0$, but it's also true that $a^2+b^2\ge 2ab \Rightarrow a^2+b^2-2ab \ge 0$. But it's only the first of these inferences that is relevant for the proof. And of course, in other proofs you might use inferences which are not reversible.
This technique is logically the same as the usual forward method, but useful when you don't know what premise to start from; instead, you take what you want to prove, and figure out what you need in order for it to be true.
My rep is too low to comment, but others are correct in saying this works because each statement is an equivalence relation. If your teacher wants to prove something by starting with an assumption, it's better to assume the opposite: $a^2+b^2<2ab$, then work through the same steps to arrive at $(a-b)^2<0$ which is a contradiction if $a$ and $b$ are real. That proves $a^2+b^2 \geq 2ab$ by contradiction.
You can not actually. You should rather start with $(a-b)^2 \geq 0$ and the show that $a^2+b^2 \geq 2ab$, not the other way round.
You can't.
By assuming first what he needed to prove, your teacher has proven $True \implies True$ which is quite useless.
The way it's written is wrong, and you're right, we can't assume something and prove that it is right.
However, instead of one way implication at each step (like $a^2+b^2\geq 2ab$ therefore $a^2+b^2 -2ab\geq 0$), we in fact have an if and only if condition. Since each statement then becomes equivalent, if you can prove any one of them you are done. And that is the case since $(a-b)^2$ is indeed greater than equal to $0$.
Contrary to what other people are saying, your teacher's proof was mostly right - it was just missing one last sentence which he might have left out: "Every step done so far is also valid in reverse". Since that is true, you can read the steps backwards and get to the premise given the conclusion. This sort of reasoning is quite common in math since the most obvious way to prove is not necessarily the way the proof will actually read.
Reading backwards gives:
We know that $(a-b)^2 \geq 0$.
$\implies a^2+b^2-2ab \geq 0$
$\implies a^2+b^2 \geq 2ab$
So it can be converted to the more "traditional" way of writing the proof.
For real numbers the square of a number is non negative. $a-b$ is a real number because $a$ and $b$ are real. Therefore $(a-b)^2$ is non negative.
The inequality $a^2+b^2\geq 2ab$ can be transformed to
$a^2-2ab+b^2\geq 0$ by substracting $2ab$ on both sides.
This is equal to $(a-b)^2$. Just multiply out the brackets.
Okay, so first take the simple case where (a == b).
Imagine two large blocks, each arranged with unit squares. On the left is A, a square with (a) rows of (a) columns. On the right is B, a square with (b) rows of (b) columns. The two squares are the same, and the total is a^2 + b^2, which is also 2ab.
Now, let's tackle the unequal case: increase (b) by exactly one. Imagine that both A and B turn into rectangles with (a) columns and (b) rows. That gives us a total of (2ab) unit squares -- but that's not what we want. We want to take the top row from block A, rotate it 90 degrees, and slap it on as the rightmost column of block B. Draw this out to see what I'm describing. Notice that now, A is again a square, with total area a^2. Block B is incomplete however: it is exactly one less than b^2.
In short, if b==(a+1), then 2ab + 1 == (a^2 + b^2). This problem gets worse as the difference between (a) and (b) increases. It follows that (a^2 + b^2) >= 2ab.
On a sidenote, if you continue to chew on this relationship, you'll discover why factoring the difference between squares as (a+b)(a-b) works, and if you play further, you can discover why F.O.I.L. works.
