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?

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    $\begingroup$ See the comment that I left for you, following the answer of Mark Saving. When writing the comment, I mis-addressed it. $\endgroup$ Jan 27, 2022 at 3:21

3 Answers 3


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.

  • $\begingroup$ Thank you. So, I understand that essentially what I should have done is taken a true statement and connect it to the claim. Which forms an implication because we know the given $P$ is true and through algebra and explanations I would show that indeed $Q$, the claim, is true. $\endgroup$ Jan 27, 2022 at 2:46
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    $\begingroup$ @TeddyMath Actually, the work that you did was very good. However, it was only half-complete. You saw that what you were trying to prove led to a true conclusion. This is the scratch work. Then, you would have to wonder whether the steps were reversible, which they are. So, after doing the scratch work, you would start with your derived equation $(x-1)^2 \geq 0.$ Then, following your own bread crumbs, you would arrive at the necessary conclusion. That is, you would take the same steps that you already took, but in reverse order. $\endgroup$ Jan 27, 2022 at 2:50
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    $\begingroup$ @user2661923: Yes - though you don't even need to start with the derived equation - you can start with: "Assume $x\in\mathbb{R}>0$. Consider the statement $x + 1/x \geq 2$", and then proceed along the same lines as in the question, verifying at each step that the implication is if and only if ($\iff$), which (in this case) it is, though only because of the assumption $x>0$. $\endgroup$
    – psmears
    Jan 27, 2022 at 10:49
  • $\begingroup$ @psmears Technically true, and also applicable in $\epsilon, \delta$ proofs around a limit. However, I personally, feel that the step by step verification of bi-conditionality is simply not worth the time/effort. Instead, once the bread crumbs are dropped, I try to validly re-trace my steps. Either the re-traced step is valid or it isn't. $\endgroup$ Jan 27, 2022 at 12:39
  • $\begingroup$ @user2661923: Obviously you're free to do proofs however works best for you - but, in cases like this at least, verifying bi-conditionality doesn't really take more time/energy - in fact, possibly less as you don't even have to rewrite all the equations ;-) $\endgroup$
    – psmears
    Jan 27, 2022 at 13:45

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.


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 $x>0$.


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