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I am trying to prove the following two inequalities using AM-GM

First, I must show that the following inequality must hold : $$ \frac{x_{1}x_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}}\leq\frac{x_{1}^{2}}{x_{1}^{2}+y_{1}^{2}}+\frac{x_{2}^{2}}{x_{2}^{2}+y_{2}^{2}} $$ and $$\frac{y_{1}y_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}}\leq\frac{y_{1}^{2}}{x_{1}^{2}+y_{1}^{2}}+\frac{y_{2}^{2}}{x_{2}^{2}+y_{2}^{2}} $$

My Attempt : Let $x_{1},x_{2},y_{1},y_{2}\in\mathbb{R}$,let $x=\sqrt{x_{1}^{2}+x_{2}^{2}}$ and $y=\sqrt{y_{1}^{2}+y_{2}^{2}}$ it appears that I could pove that $\sqrt{\sum_{i=1}^{2}x_{i}^{2}y_{i}^{2}}\leq xy$ as follows : $$ xy=\left(\sqrt{x_{1}^{2}+x_{2}^{2}}\right)\left(\sqrt{y_{1}^{2}+y_{2}^{2}}\right)=\sqrt{x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}}\geq \sqrt{x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}} $$ I noticed that it implies that $\frac{\sqrt{x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}}}{\left(\sqrt{x_{1}^{2}+x_{2}^{2}}\right)\left(\sqrt{y_{1}^{2}+y_{2}^{2}}\right)}\leq 1$

I am unable to continue I hope someone gives me a hint.

  • Does this answer your question? Proving Cauchy-Schwarz with Arithmetic Geometric mean – Martin R Oct 04 '20 at 13:16
  • Dear @MartinR, my answer would yes if and only if someone can show me how this inequality is derived : $\sum_{i=1}^n \frac{a_ib_i}{AB} \leq \sum_{i=1}^n \frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right) = 1$ from the user's question –  Oct 04 '20 at 13:20
  • I'm thinking that your first 2 inequalities are not the form of Cauchy-Schwarz, as I plugged $1$ in $x_1, x_2, y_1, y_2$ and the equality did not happen. Where did you take those inequalities from? – Đào Minh Dũng Oct 04 '20 at 13:21
  • @WiWo: That is the AM-GM inequality, as pointed out in the answer https://math.stackexchange.com/a/2925717/42969. – Martin R Oct 04 '20 at 13:21
  • @MartinR but I only know AM-GM of the form $\frac{a_{1}+...+a_{n}}{n}\geq\sqrt[n]{a_{1}\cdot...\cdot a_{n}}$, how could you prove that it also has the form you mentioned. –  Oct 04 '20 at 13:29
  • @WiWo: $\frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right) \ge \sqrt{ \frac{a_i^2}{A^2} \cdot \frac{b_i^2}{B^2} } = \frac{|a_i b_i|}{AB}$, using AM-GM for two numbers. – Martin R Oct 04 '20 at 13:32
  • I understand, but is $\frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right)$ assumed in its form like this or derived somehwere? that is to say did they say assume $x=\frac{a_{1}^{2}}{A^{2}}$ and $y=\frac{b_{1}^{2}}{B^{2}}$ or is it derived from an inequality? and this would answer my question. –  Oct 04 '20 at 13:41
  • @ĐàoMinhDũng If you divide the RHS by 2, then you get a stronger inequality which follows from C-S – Omer Oct 04 '20 at 13:42
  • @Omer indeed, I just feel surprised that the $\frac 1 2$ is not included in the inequality. Maybe he just missed adding it – Đào Minh Dũng Oct 04 '20 at 13:47
  • Is there such need? because $2\sqrt{xy}>\sqrt{xy}$ –  Oct 04 '20 at 13:50
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    @WiWo If you're trying to deduce C-S from those inequalities, then yes, you need it. Because after you sum those inequalities you want to get $1$ on the RHS and not $2$. – Omer Oct 04 '20 at 13:56
  • @Omer you are correct, I just obtained 2 instead of 1. :( –  Oct 04 '20 at 14:45

2 Answers2

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We can even prove Cauchy-Schwartz using AM-GM for $\mathbb{R^n}$. Note that by AM-GM:
$$2 = 1+1 = \sum_{i=1}^{n} \frac{a_i^2}{a_1^2+...+a_n^2} + \sum_{i=1}^{n} \frac{b_i^2}{b_1^2+...+b_n^2} = \sum_{i=1}^{n} \left(\frac{a_i^2}{a_1^2+...+a_n^2} + \frac{b_i^2}{b_1^2+...+b_n^2}\right) \geq \sum_{i=1}^{n} \frac{2a_ib_i}{\sqrt{(a_1^2+...+a_n^2)(b_1^2+...+b_n^2)}}$$.
Dividing both sides by $2$ and squaring both sides gives us the desired inequality.

For your inequality, if we let $x = \frac{x_1^2}{x_1^2+y_1^2}, y = \frac{x_2^2}{x_2^2+y_2^2}$, Then we are trying to prove $x+y \geq \sqrt{xy}$, Which follows immediatly from AM-GM as $x+y \geq 2\sqrt{xy}$ (and we see that the original inequality will be true even if we divide the RHS by $2$).

Omer
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  • It appears to be different from my expression as I must have on the numerator $a_{1}a_{2}$ for instance instead of $a_{1}b_{1}$. Maybe I must consider redefining in my question $x$ as $x=\sqrt{x_{1}^{2}+y_{1}^{2}}$ perhaps? –  Oct 04 '20 at 13:17
  • If I understand correctly, you can assume $x_1,x_2,y_1,y_2 \geq 0$, and then if we let $x = x_1^2/(x_1^2+y_1^2), y = x_2^2/(x_2^2+y_2^2)$, then what you are actually trying to prove is $x+y \geq \sqrt{xy}$, which follows immediatly from AM-GM (or by squaring both sides) as $x+y \geq 2\sqrt{xy}$. Thus the original inequality will hold even if you divide the right side by $2$. – Omer Oct 04 '20 at 13:32
  • I believe you have made things a lot clear. However, can we conclude the expression $x_1^2/(x_1^2+y_1^2), y = x_2^2/(x_2^2+y_2^2)$ somewhere from an inequality? instead of assuming they exist in this form? –  Oct 04 '20 at 13:38
  • I don't fully understand what you're asking here. Can you be more specific? – Omer Oct 04 '20 at 13:43
  • Ignore my previous question I believe you have fully answered my question, you can type your answer as the one you mentioned in your comment which is about $x+y>\sqrt{xy}$ and I will mark it as a correct answer if you wish. –  Oct 04 '20 at 13:45
  • @WiWo I have edited my answer. Hopefully it fully answers your question. – Omer Oct 04 '20 at 13:54
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So assuming that $x$ is a vector with coordination $(x_1, x_2)$ and $y$ is a vector with coordination $(y_1, y_2)$, then the length of $x$ is $|x| = \sqrt{x_1^2 + x_2^2}$, and the length of $y$ is $|y| = \sqrt{y_1^2 + y_2^2}$. Take the dot product of 2 vectors $x$ and $y$ you will have $x\cdot y = x_1y_1 + x_2y_2$

Noticing that the definition of dot product is $x \cdot y = |x||y|\cos\alpha$ where $\alpha$ is the angle between vector $x$ and $y$, we will have $$\frac{x\cdot y}{|x||y|} = \cos \alpha \leq 1$$ $$\Rightarrow \frac{x_1y_1 + x_2y_2}{\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}} \leq 1$$ and we are done

Đào Minh Dũng
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  • I unfortunately can not use vector product or vectors in general. I am restricted to use AM-GM inequality I wanted to show that $\frac{x_{1}x_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}}\leq\frac{x_{1}^{2}}{x_{1}^{2}+y_{1}^{2}}+\frac{x_{2}^{2}}{x_{2}^{2}+y_{2}^{2}}\quad\text{and}\quad\frac{y_{1}y_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}}\leq\frac{y_{1}^{2}}{x_{1}^{2}+y_{1}^{2}}+\frac{y_{2}^{2}}{x_{2}^{2}+y_{2}^{2}}$ from which I can deduce the Cauchy-Schwarz –  Oct 04 '20 at 12:57
  • @WiWo sure I can give another answer. Hint: you can square both sides of the inequality $\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2} \leq x_1y_1 + x_2y_2$. I will give an answer later for you to read – Đào Minh Dũng Oct 04 '20 at 13:01
  • thank you, but how did you come up with this inequality in the comment section. –  Oct 04 '20 at 13:04
  • if by chance, other than vector dot product. –  Oct 04 '20 at 13:05
  • @WiWo so what is your original inequality? I did not see that in your question, only your attempt – Đào Minh Dũng Oct 04 '20 at 13:05
  • you are correct, I will edit my question to make things clearer. –  Oct 04 '20 at 13:07