I'm studying linear algebra. I'm a beginner in this subject. The book says: Cosine formula of two vectors $$\frac{v \cdot w}{||v|| ||w||} = cos\theta $$

and it says, since $|cos\theta|$ never exceeds 1, the cosine formula gives two great inequalities:

$$Schwarz Inequality : |v\cdot w| \le ||v|| ||w||$$ $$Triangle Inequality : ||v+w|| \le ||v||+||w||$$

my questions are:

can I derive Schwarz Inequality from the Cosine formula of two vector which the process will be:

$$\frac{v \cdot w}{||v|| ||w||} = cos\theta $$ this means $$ -1 \le \frac{v \cdot w}{||v|| ||w||} \le 1 $$ because $\cos \theta$ is between -1 and 1. Then I take only the absolute of $ \frac{v \cdot w}{||v|| ||w||} $ I will get $$0 \le | \frac{v \cdot w}{||v|| ||w||} | \le 1$$ which means: $$0 \le | v \cdot w|*|\frac{1}{||v|| ||w||}| \le 1$$ Then I multiply all by $$0 * ||v||||w|| \le | v \cdot w|*|\frac{1}{||v|| ||w||}| *||v||||w|| \le 1 *||v||||w||$$

Then I get $$0\le |v \cdot w| \le ||v||||w||$$

Which is the Schwarz Inequality and can be read as "The absolute value of dot product between v and w is greater than or equal to zero and less than or equal length of v times length of w"

Is all my work correct? and is it legitimate to derive Schwarz Inequality in this way?

My second question is how to derive Triangle inequality from the cosine formula at the top ? I know triangle inequality from college algebra but have no idea how to derive it from the cosine formula on the top (as it said in the book).

  • 1
    $\begingroup$ But how would you get that cosine equation without Cauchy-Schwarz? $\endgroup$ Apr 28, 2016 at 6:22
  • $\begingroup$ if you have two unit vector cos(a), sin(a) and cos(b), sin(b) the dot product of this two vector gives you cos(b-a) (or in other word the cos of angle between them) then if you have any vector and you make it to be unit vector you can use this formula. and this is always true without knowing Cauchy-Schwarz. I never heard of Cauchy-Schwarz formula because I'm newbie but I still can prove Cosine formula of two vectors. but can the cosine formula be used as one way to get into Cauchy-Schwarz formula ? as it is said in the book $\endgroup$ Apr 28, 2016 at 6:28
  • $\begingroup$ Were you ever able to get your answer? I have the same question. I am reading a book and one of the exercises asks to explain why deriving C-S inequality this way is invalid. Which means, you can't derive it this way. $\endgroup$
    – Sun
    Jul 3, 2018 at 1:04
  • $\begingroup$ I believe @RickSanchez answer is what you are looking for. $\endgroup$
    – Sun
    Jul 3, 2018 at 2:01

2 Answers 2


Yes, your work is correct, although the leftmost inequality is redundant, an absolute value is always nonnegative, you don't have to keep writing it. To derive the triangle inequality from CS, square both sides first $$\|v+w\|^2\leq(\|v\|+\|w\|)^2$$ and notice that the LHS is equal to $$(v+w)\cdot(v+w)=v\cdot v+v\cdot w+ w\cdot v+w\cdot w=\|v\|^2+2v\cdot w+\|w\|^2.$$ Expand the RHS as well and cancel out like terms to arrive at $$v\cdot w\leq\|v\|\|w\|.$$ This holds true so $$\|v+w\|^2\leq(\|v\|+\|w\|)^2$$ is also true. Taking square roots yields the triangle inequality.

  • $\begingroup$ where is |v dot w| ? I only see v dot w ? dose square root at the beginning generate |v dot w| ?? $\endgroup$ Apr 28, 2016 at 6:40
  • 2
    $\begingroup$ There is no square root at the beginning. You assume that $|v\cdot w|\leq\|v\|\|w\|$, but that clearly implies $v\cdot w\leq\|v\|\|w\|$, as $x\leq|x|$ for any real number $x.$ $\endgroup$
    – ε-δ
    Apr 28, 2016 at 6:45
  • $\begingroup$ great, please change the u to v. I think there is an error. where u comes from ? $\endgroup$ Apr 28, 2016 at 6:50
  • 1
    $\begingroup$ That was just a typo. Fixed. $\endgroup$
    – ε-δ
    Apr 28, 2016 at 7:00
  • $\begingroup$ one question, when you write $||v+w||^2$ is it equal to dot product of v+w with it self or (v+w)*(v+w) is it result in the same constant if I do either way? $\endgroup$ Apr 28, 2016 at 9:05

It's just around the other way: because of the CS-inequality one can define the cosine between two vectors ...

  • $\begingroup$ I know that the CS-inequality can help me find cosine between two vectors but can I use it to derive the Schwarz Inequality as it is said in the book? and other question. please answer all my question above. $\endgroup$ Apr 28, 2016 at 6:16

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