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I'm trying to prove this but am having some difficulty.

For any $x,y\in\mathbb R$ such that $x\ge 0$ and $y\ge 0$ we have $$\frac{x+y}2 \ge \sqrt{xy}.$$

So far what I have gotten to is $\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2} $

After this point I don't know what to do. To get to this point:

$$y \geq 0 $$ $$\implies y \geq y$$ $$\implies xy \geq xy$$ $$\implies 2xy \geq xy$$ $$\implies x^{2} + 2xy + y^{2} \geq xy$$ $$\implies (x+y)^{2} \geq xy$$

Sqrt both sides to get:

$$\implies x+y \geq \sqrt{xy}$$ $$\implies\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2}$$

Martin Sleziak
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Leo Denni
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  • How did you get to that point? (I actually don't see a way to get there without solving the problem all the way, so I'm curious.) – Noah Schweber Jan 30 '16 at 04:18
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    I updated the post to show my work – Leo Denni Jan 30 '16 at 04:26
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    Why have you deleted your work? I think it was beneficial to the question. – YoTengoUnLCD Jan 30 '16 at 06:14
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    The work leads to a dead end, so I feel it will mislead anyone else reading it. – Leo Denni Jan 30 '16 at 06:18
  • See also http://math.stackexchange.com/questions/64881/proving-the-am-gm-inequality-for-2-numbers http://math.stackexchange.com/questions/904827/how-to-prove-that-fracab2-geq-sqrtab-for-a-b0 http://math.stackexchange.com/questions/1114615/if-0ab-prove-that-a-sqrtab-fracab2b http://math.stackexchange.com/questions/1150895/let-a0-and-b0-prove-that-sqrtab-le-ab-2 http://math.stackexchange.com/questions/543253/how-can-i-prove-frac2xyxy-leq-sqrtxy-leq-fracxy2 – Martin Sleziak Jan 30 '16 at 09:46
  • And if you denote $x=a^2$ and $y=b^2$, then you can look at http://math.stackexchange.com/questions/241741/simple-algebra-question-proving-a2b2-geqslant-2ab http://math.stackexchange.com/questions/320244/show-that-2-xy-x2-y2-for-x-is-not-equal-to-y http://math.stackexchange.com/questions/470221/prove-the-inequality-xy-leq-frac12x2y2 http://math.stackexchange.com/questions/943994/show-that-for-all-real-numbers-a-and-b-ab-le-1-2a2b2 – Martin Sleziak Jan 30 '16 at 09:47
  • BTW you can use $\implies$ $\implies$ or $\Rightarrow$ $\Rightarrow$. It looks a bit better than $=>$. – Martin Sleziak Jan 30 '16 at 09:50
  • @LeoDenni Even if your work would prove to be misleading if we read it, we are here to help you and identify any errors in your working out to verify whether or not you have proved the inequality. From the looks of it, you have errors, and so we can now most certainly help you out. That is what the MSE is for :) – Mr Pie Feb 01 '18 at 09:23
  • Here is another link discussing the matter $\longrightarrow$ https://math.stackexchange.com/questions/691807/proofs-of-am-gm-inequality There are also other links accessible from the post. – Mr Pie Feb 01 '18 at 09:30

3 Answers3

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$$(x-y)^2 \ge 0\\ x^2+2xy+y^2-4xy \ge 0 \\(x+y)^2 \ge 4xy\\\frac {x+y}2 \ge \sqrt{xy}$$

Ross Millikan
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  • I have no idea what is going on here – Leo Denni Jan 30 '16 at 04:33
  • Any square is greater than zero, for the first line. I added a step. From the third to fourth, take the square root – Ross Millikan Jan 30 '16 at 04:39
  • Well first of all, $(x-y)^{2} \neq x^{2} + 2xy + y^{2}$. And also how can you subtract $4xy$ from the equation in the second line without giving any reason? How do I know the inequality is still true? – Leo Denni Jan 30 '16 at 04:42
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    @LeoDenni $$x^2+2xy+y^2-4xy=x^2-2xy+y^2=(x-y)^2$$ –  Jan 30 '16 at 04:52
  • from the second to the third I subtracted $4xy$ from both sides and noted that $x^2+2xy+y^2=(x+y)^2$ – Ross Millikan Jan 30 '16 at 04:53
  • I kind of get it, but I'm still confused. If I have $x^{2} + 2xy + y^{2}$, how am I able to subtract $-4xy$ from that and keep the inequality true? It just seems like it is out of no where – Leo Denni Jan 30 '16 at 04:58
  • @LeoDenni The point is that $x^2+2xy+y^2-4xy$ is the same as $x^2-2xy+y^2$ - this is because $2xy-4xy=-2xy$. Really there's steps in between - $(x-y)^2=x^2-2xy+y^2=x^2-2xy+y^2+4xy-4xy=x^2+2xy+y^2-2xy$. – Noah Schweber Jan 30 '16 at 05:01
  • Okay I understand now, but I'm not satisfied. I just don't see how you knew to make that jump. How was I suppose to know to do that? (Please bear in mind I'm very new to proofs) – Leo Denni Jan 30 '16 at 05:08
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    Understanding a proof is very different from finding it. To understand it, you just need to follow through step by step, filling in as necessary to make sure you can see where each comes from. I've seen this one before, so finding it is easy. The fact that $(x+y)^2-(x-y)^2=4xy$ comes up with some frequency, so if I hadn't seen it before I would have been tempted to take what you have to prove, multiply by $2$ to clear the fraction and square to get rid of the root. The more proofs you read, the more you get a feel for them, which makes finding them much easier. – Ross Millikan Jan 30 '16 at 05:20
  • Thanks for the insight. Based on my work in my original post, is there a way for me to have proven it from my last line? Or that was a dead end? – Leo Denni Jan 30 '16 at 05:25
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    That was a dead end because your RHS is too small. Just putting in the factor $2$ in the fourth line is too poor an approximation. If $x$ is close to $y$ the inequality you are trying to prove is rather close. A different proof starts out with $(\sqrt x - \sqrt y)^2 \ge 0$. You might see if you can get from there to the end. – Ross Millikan Jan 30 '16 at 05:36
  • I know that this is more than an year old, but, I feel like one of Leo Denni's questions was not answered properly. To reproduce this proof, think about it forwards and then write it formally. Start with the statement you wish to prove, manipulate it into a true statement, and if you can, go from the true statement to the statement you wish to prove. If you see that you can't write the proof backwards, then that is a red flag to check your proof for errors. – Nairit Jul 30 '17 at 18:41
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Let us start from the beginning and simplify what we want to show:

$$\frac{x+y}{2}\geq \sqrt{xy}$$

By multiplying both side by $2$, we will get:

$$x+y\geq 2\sqrt{xy}$$

Now, subtracting $2\sqrt{xy}$ from both sides, we left with:

$$x-2\sqrt{xy}+y\geq 0$$

Now, a question:

Why $x-2\sqrt{xy}-y$ is indeed greater(or equal) than zero?

Closer look on the above expression, one might see that its equal to $(\sqrt{x}-\sqrt{y})^2$

So, if we are given two positive real numbers $x$ and $y$, then without loss of generality we can say that $x>y$(why?), therefore $\sqrt{x}>\sqrt{y}$(prove that). Hence: $$\sqrt{x}-\sqrt{y}>0$$

And...

Salech Alhasov
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This is a comment on your attempt to prove this inequality rather than answer to your question. Just to show that it was not completely dead end, as you wrote in your comment.

You wrote that you tried this:

So far what I have gotten to is $\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2} $

After this point I don't know what to do. To get to this point:

$$y \geq 0 $$ $$\implies y \geq y$$ $$\implies xy \geq xy$$ $$\implies 2xy \geq xy$$ $$\implies x^{2} + 2xy + y^{2} \geq xy$$ $$\implies (x+y)^{2} \geq xy$$

Sqrt both sides to get:

$$\implies x+y \geq \sqrt{xy}$$ $$\implies\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2}$$

What you could do know could be ask yourself: Well, I want to prove stronger inequality. Could I get it by similar approach?

So we start from the end. You want to get $$\frac{x+y}{2} \geq \sqrt{xy}$$ which is the same as $$x+y \geq 2\sqrt{xy}.$$ Relation between these two is the same as in the last step of your attempt - just divided by $2$. If we try to go one step back in the same way as above, this can be obtained as the square root of $$ \begin{align*} (x+y)^2 &\ge 4xy\\ x^2+2xy+y^2 &\ge 4xy\\ \end{align*} $$

So at this point you could think about: "How could I get $x^2+2xy+y^2 \ge 4xy$?" Maybe after playing a bit with this expression you would found a way to prove this. (However, now you already know how to show this from other answers.)

Community
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Martin Sleziak
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  • And since I have already posted so many links in comments, maybe it is ok to post a few more - other posts about the weaker inequality, which you have shown in your attempt (to be more precise, about some equivalent or very similar inequality): http://math.stackexchange.com/questions/357272/how-can-i-prove-that-xy-leq-x2y2, http://math.stackexchange.com/questions/920605/inequality-x2y2xy-ge-0, http://math.stackexchange.com/questions/1439494/prove-that-a2abb2-ge-0 – Martin Sleziak Jan 30 '16 at 10:13