How to know when I've proved a claim (with example)

Question

I have a homework problem and want to try to understand the process without following steps from other outside sources. I understand basic logic and have done a handful of introductory proofs to this point. My problem is, when I am doing my scratchwork and gathering evidence, sometimes I can feel like I've just gone full circle and not addressed the claim or that I've completed the proof but am unaware of it.

For this problem, it asked me to prove that: For all positive real numbers $x$, the sum of $x$ and its reciprocal is greater than or equal to $2$.

This is my work currently:

Suppose $x + \frac{1}{x} \geq 2$ where $x \in\mathbb{R}$ and $x > 0$.

Multiplying by $x$, we obtain $x(x+\frac{1}{x})\geq2x$

And $x(x+\frac{1}{x})=x^{2}+1\geq2x$

Subtracting $2x$ from both sides gives $x^{2}-2x+1\geq0$

And further we see $x^{2}-2x+1=(x-1)^{2}\geq0$ which is true for any real number $x$.

Thus, it has been shown. $\square$

Does this count as a proof? I'm not sure if my assumption is the correct way to go about it. I understood that you assume $P$ and show that it means $Q$ is true. But in this case it feels like I'm assuming $P\rightarrow Q$ which doesn't make sense to me.

For a clear question, could someone explain to me the logical groundwork behind a proof like this? What constitutes proven and why does whatever is shown justify the claim to be proven?

Answer 1

No, this is not a proof. You started with the claim $x + 1/x \geq 2$ and derived a true statement. To provide a proof, you must start with known true statements and derive the fact that $x + 1/x \geq 2$.

Essentially, you did the proof backwards.

The correct proof would go as follows:

Note that $x^2 - 2x + 1 = (x - 1)^2 \geq 0$. Adding $2x$ to both sides, we see that $x^2 + 1 \geq 2x$. Now since $x$ is positive, we can divide both sides by $x$ to see that $x + 1/x \geq 2$, as required.

If you really want to know that you have provided a complete proof, you would translate this into a formal proof and provide a rigorous justification for every step.

Answer 2

I agree with the answer of Mark Saving. An alternate (valid) approach would be a proof by contradiction.

That is, you would start with the premise that

$$x + \frac{1}{x} < 2 ~: ~x \in \Bbb{R^+}. \tag1 $$

Then, following virtually the exact same steps that you took, you would conclude that

$$(x - 1)^2 < 0. \tag2 $$

Then, you would reason that (2) above is impossible. Therefore, since (2) above was implied by (1) above, (1) above is impossible.

Answer 3

Yes, you are assuming what needs to proven as you wrote "Suppose that $x + \frac{1}{x} \geq 2$". As an alternative proof to Mark Saving's answer, you can use AM-GM:

$$\frac{\displaystyle x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} \\ \frac{x + \frac{1}{x}}{2} \geq 1 \\ \displaystyle x + \frac{1}{x} \geq 2.$$ This inequality holds for all $x > 0$. Therefore, $x + \frac{1}{x} \geq 2$ for $x>0$.