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First and foremost, I have already gone through the following posts:

Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$

Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$

The reason why I open a new question is because I do not understand after reading the two posts.

Question: Prove that for any two positive numbers x and y, $\sqrt{ xy} \leq \frac{x + y}{2}$

According to my lecturer, he said that the question should begin with $(\sqrt{x}- \sqrt{y})^2 \geq 0$. Lecturer also said that this is from a "well-known" fact. Now, both posts also mentioned this exact same thing in the helpful answers.

My question is this - how and why do I know that I need to use $(\sqrt{x}- \sqrt{y})^2 \geq 0$? What "well-known" fact is this? Can't I simply just subtract $\sqrt{xy}$ to both side and conclude at $0 \leq {(x-y)}^2$? I do not know how this $(\sqrt{x}- \sqrt{y})^2 \geq 0$ come back and why it even appear.

Thanks in advance.

Edit: I am not looking for the direct answer to this question. I am looking for an answer on why $(\sqrt{x}- \sqrt{y})^2 \geq 0$ is even considered in the first place as the first step to this question. Is this from a mathematical theorem or axiom etc?

SunnyBoiz
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    The square of any real number is $\geq0$ – saulspatz May 12 '20 at 12:55
  • Thanks. I understand your statement. But why involve $\sqrt{x}$ and $\sqrt{y}$ in the equation? Where does the square root come out from? – SunnyBoiz May 12 '20 at 12:56
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    Multiply it out and you'll see. – saulspatz May 12 '20 at 12:57
  • may be useful for you: https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means – Abdullah AKKURT May 12 '20 at 12:57
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    @saulspatz I don't feel like you are really explaining to me in details. I know that multiplying out will get back the result which prove the statement. My question is why do I even know that I must use this in the first place? What is this "well-known" fact? Is it a theorem? – SunnyBoiz May 12 '20 at 12:59
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    The “well-known fact” is that the square of a real number is non-negative. It is applied to $(\sqrt{x}- \sqrt{y})^2$ because that leads to the wanted result. – Martin R May 12 '20 at 13:04
  • @MartinR The reason why square root is involved is so that once I expand it out, I can get back the same statement as per the question? Is this the reason why it is being used as this can prove that for all positive x and y, it is true? (I know the square part). – SunnyBoiz May 12 '20 at 13:06
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    @SunnyBoiz "why do I even know that I must use this in the first place?" , It is not a known fact, just a fact that if you know it and you use it then it will take you to what you want to show. What really happens in practice is this, they start with end result and try to work backwards, until they get to something like $\sqrt{xy}= \frac{x+y}{2}$, now if they can show that is true then it all works, once you know it is true then you write the proof backwards and you look smart! just try it, write the proof backwards and you see why $\sqrt{xy}= \frac{x+y}{2}$ is true then proof! – jimjim May 12 '20 at 13:28

5 Answers5

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We would like to prove $$ \frac{{x + y}}{2} \ge \sqrt {xy} $$ for all non-negative $x$, $y$. If this was true, then we would also have $$ x + y \ge 2\sqrt {xy}, $$ $$ x - 2\sqrt {xy} + y \ge 0, $$ $$ \left( {\sqrt x - \sqrt y } \right)^2 \ge 0. $$ But this last inequality is certainly true for all non-negative $x$, $y$. This is because the square of any real number is non-negative. Now you can start with this "well-known fact" and do everything backwards to obtain $\frac{{x + y}}{2} \ge \sqrt {xy} $. This shows you how one might come up with the idea of starting with $\left( {\sqrt x - \sqrt y } \right)^2 \ge 0$. This is a common proof technique in mathematics when you have a series of equivalent transformations that can be performed in both directions.

Gary
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  • Hi there. It looks like I will need to understand and do more mathematics to see that ${(\sqrt{x}-\sqrt{y})}^2 \geq 0$ will yield back the same result. I think I understand now why everyone suggested to do this first step - that is because everyone knows that it is a positive x and y (thus squaring) and with the square root will get the same statement. This is also a lot easier to prove (if you can see it and deduce the first step immediately). My original answer was the same as Narasimham in this post. Is his method actually also applicable? That is, there are more than one way to prove this? – SunnyBoiz May 12 '20 at 14:32
  • To see that $\left( \sqrt x - \sqrt y \right)^2 \ge 0$ yields the original inequality just read my derivation from the bottom upwards. Narasimham's proof works as well. – Gary May 12 '20 at 14:54
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This proof is hinted by the presence of the square root, which one will tend to remove by squaring. As all numbers are positive

$$\sqrt{xy}\le\frac{x+y}2$$ is rewritten

$$xy\le\frac{x^2+2xy+y^2}4,$$

which is also

$$0\le\frac{x^2-2xy+y^2}4$$ and certainly holds (see why ?).

Once you have understood this principle, you can recast the proof as

$$\sqrt x\sqrt y\le\frac{\sqrt x^2+\sqrt y^2}2$$ or $$0\le(\sqrt x-\sqrt y)^2.$$

An alternative way to eliminate the square root is to set, like in Narasimham's answer, $x=u^2,y=v^2$ and prove

$$uv\le\frac{u^2+v^2}2.$$

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Let $ x=u^2, y=v^2$ since $ (x,y)$ are given positive.

You have to prove that

$$ u^2+v^2-2 u v \ge 0$$

or that

$$(u-v)^2\ge 0 $$

which is true for all real numbers.

Narasimham
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The well-known fact your teacher is talking about is that the square of any really number is nonnegative. That is a theorem. As for why you should start with this particular instance of that theorem, that takes a little insight.

If you were trying to prove the theorem yourself, you might start by seeing what happens if it isn't true. $$\sqrt{xy}>\frac{x+y}2\\ xy>\frac{x^2+2xy+y^2}4\\ 4xy>x^2+2xy+y^2\\ 0>x^2-2xy+y^2\\ 0>(x-y)^2$$
contradiction.

This is a perfectly good proof, but you might prefer a direct proof. If you try to work backwards, you immediately run into trouble, because taking the square root of both sides may not be valid. It may then occur to you to start with the square roots.

saulspatz
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$$\sqrt{xy} \leq \frac{x+y}{2}$$ $$xy \leq \frac{(x+y)^2}{4}$$ $$4xy \leq x^2+2xy+y^2$$ $$2xy \leq x^2+y^2$$ $$2 \leq \frac{x^2}{xy}+\frac{y^2}{xy}$$ $$2 \leq \frac{x}{y}+\frac{y}{x}$$ Since $\frac{x}{y}$ is a fraction, assume $x$ is greater than $y$, say $$z = \frac{x}{y}$$ This means that the lowest value of $z$ is $1$, if $x > y$ $$2 \leq z + \frac{1}{z}$$ If $z$ is lowest, $\frac{1}{z}$ would be highest, but the minimum value of their sum comes when they're minimum.....equal to $1$ $$z + \frac{1}{z} ≥ 2$$

Gary
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Aderinsola Joshua
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