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I am having trouble with this problem from my latest homework.

Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x,y$, we have $$ \sqrt{xy}≤ \frac{x+y}{2} .$$ Furthermore, equality occurs if and only if $x = y$.

Any and all help would be appreciated.

Martin Sleziak
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Steve
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    One way is the following. Let $\sqrt{x} = a$ and $\sqrt{y} = b$. Substitute for $x$ and $y$ in terms of $a$ and $b$. Collect all the terms together on the right side, and factor. Do you recognize a familiar inequality? – Srivatsan Sep 15 '11 at 19:51

5 Answers5

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Since $x$ and $y$ are positive, we can write them as $x=u^2$, $y=v^2$. Then

$$(u-v)^2 \geq 0 \Rightarrow u^2 + v^2 \geq 2uv$$

which is precisely it.

Bruno Joyal
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Note that $$\frac{x+y}{2}-\sqrt{xy}=\frac{(\sqrt{x}-\sqrt{y})^2}{2}.$$

André Nicolas
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I am surprised no one has given the following very straightforward algebraic argument: \begin{align} 0\leq(x-y)^2&\Longleftrightarrow 0\leq x^2-2xy+y^2\tag{expand}\\[0.5em] &\Longleftrightarrow 4xy\leq x^2+2xy+y^2\tag{add $4xy$ to both sides}\\[0.5em] &\Longleftrightarrow xy\leq\left(\frac{x+y}{2}\right)^2\tag{div. sides by 4 & factor}\\[0.5em] &\Longleftrightarrow \sqrt{xy}\leq\frac{x+y}{2}.\tag{since $x,y\in\mathbb{R}^+$} \end{align} In regards to equality, notice that $\sqrt{xy}\leq\frac{x+y}{2}\leftrightarrow 2\sqrt{xy}\leq x+y$, and it becomes clear that equality holds if and only if $x=y$. $\blacksquare$

Daniel W. Farlow
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  • I have, in a duplicate of this question, see https://math.stackexchange.com/a/906130/11994. As I explain there, I write this proof up-side-down, to reduce surprises for the reader, by making it essentially a simplification proof. – MarnixKlooster ReinstateMonica Jan 15 '18 at 05:25
  • @MarnixKlooster You can do that, but it's often considered bad form (and oftentimes introduces logical errors). See this write up from a Stanford prof about proof writing--what you did was mistake #1. In my proof, I start with the common knowledge that the square of any real number is nonnegative from which I work out the rest of the argument. In fact in my argument above, the last $\iff$ should be a $\implies$. Oh well. Such is life. But just know you always want to work towards the conclusion and not the other way round, even to simplify – Daniel W. Farlow Jan 15 '18 at 21:32
  • I take the liberty to disagree here. :-) Presenting a proof to follow the intuition that led you to the proof, and thus removing unnecessary surprises ("rabbits") for the reader, makes a proof easier to read. See Dijkstra's EWD1300. What do you think of the readability and clarity of your proof presentation vs mine? – MarnixKlooster ReinstateMonica Jan 18 '18 at 23:09
  • By the way, this probably belongs elsewhere, perhaps in a new question on proof presentation. Perhaps I could quote your comment above in such a question, for concreteness? – MarnixKlooster ReinstateMonica Jan 18 '18 at 23:10
  • @MarnixKlooster I don't think we are necessarily "disagreeing" here. Your response to mine had to do with readability and clarity, something I said nothing about in my comment. My comment had to do with logical correctness and proof presentation. In that regard, it does not make logical sense to work from something you are trying to prove to something you already know. As that linked write up indicates, do whatever you want in the process of figuring something out, but final presentation should be polished...even if it's "pedagogically broken." Rudin is a great example of what I mean. [...] – Daniel W. Farlow Jan 19 '18 at 02:00
  • [...] His POMA book is logically beautiful. Clean and precise. But it's pedagogically horrific. There's no motivation for anything. Its readability is questionable at best (for beginners at least) albeit the clarity becomes something to be treasured after one has wrestled with it for some time. I remember reading in a book by Hubbard how Jacobi would complain that Gauss' proofs often seemed unmotivated to which Gauss reportedly said, "You build the building and remove the scaffolding." Ideally, if we math people were less lazy perhaps, we would provide a clean, logically precise proof. [...] – Daniel W. Farlow Jan 19 '18 at 02:03
  • [...] And then we would follow that up with exposition about the motivation for it and how things tie together and so on and so forth. Take construction of the reals for instance. Pretty much any proof will be a tedious affair. The one given in Rudin's POMA (using cuts) is clean and precise but completely lacking in any sort of real motivation (not in terms of the need for a proof but in terms of why the proof is structured the way it is). Any experienced mathematician who wants a refresher on such a construction would arguably not want a fluffy proof with motivation. Just a clean proof. – Daniel W. Farlow Jan 19 '18 at 02:07
  • Perhaps what all I just wrote is not as articulate as it could be, but I don't think we are disagreeing on things so much as we are trying to emphasize different things. I am trying to emphasize logical correctness (after all, we are dealing with a proof) whereas it seems you are strongly emphasizing readability and clarity. Ideally you would have both, but that's exceedingly difficult to accomplish. Anyway, feel free to quote whatever you like, but I think it's worth contemplating the fact that I think we agree more than we disagree. – Daniel W. Farlow Jan 19 '18 at 02:09
  • Thanks for your extensive reply! I think we do disagree :-) and would value your opinion on EWD1300. Meanwhile I'll try to formulate a question about this topic-- but it might take me a little while. – MarnixKlooster ReinstateMonica Jan 19 '18 at 05:27
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$$0\le ({\sqrt x}-{\sqrt y})^{2}$$ $$0\le x-2{\sqrt {xy}}+y$$ $$2{\sqrt {xy}}\le x+y$$ $${\sqrt {xy}}\le {x+y\over2}$$

Mongol-genius
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$\phantom{Proof without words.........}$ enter image description here

lhf
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