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Let $a$, $b$, and $c$ be vectors. How do we prove that $|a + b + c| \leq |a| + |b| + |c|$?

I was able to prove that $|a + b + c| < |a| + |b| + |c|$. How do I complete the proof?

OP's work on the problem

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    Say $a_1 = -1, b_1 = 2$, and $c_1 = 3$, then $|a_1 + b_1 + c_1| = 4$, whereas $|a_1| + |b_1| + |c_1| = 6$. Contradiction. For simplicity, here, I've just used the first element of each vector. – 2017-01-10
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    Its a little bit confusing to me.Would you mind explaining it again? Thank you – 2017-01-10
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    Which part exactly is confusing? – 2017-01-10
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    @nbro that 4$ and 6$ part? And the way to prove that L.H.S=R.H.S with that? – 2017-01-10
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    The statement (or equation) can either be true or false. In order for a statement to be true, it must be true for all cases, i.e., for all vectors $a, b$ and $c$. I picked $3$ vectors $a, b$ and $c$, whose first elements summed together show us that the statement is not true. In other words, I assumed that $|a + b + c| = |a| + |b| + |c|$, but then I found that the three mentioned vectors make $|a + b + c| = |a| + |b| + |c|$ not true and thus false, so $|a + b + c| \neq |a| + |b| + |c|$, in general. – 2017-01-10
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    The $4$ comes from the fact that $|-1 + 2 + 3| = |-1 + 5| = |4| = 4$. The 6 comes from $|-1| + |2| + |3| = 1 + 2 + 3$. – 2017-01-10
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    In other words, you cannot prove it because it is not true (unless some additional restrictions are put on $a, b, c$). – 2017-01-10
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    The problem appears to be: 1. The original (assigned?) question asked for a proof that if $a$, $b$, and $c$ are vectors, then $|a + b + c| \leq |a| + |b| + |c|$. 2. This got broken into two conditions, "$<$ _and_ $=$", when it logically meant "$<$ _or_ $=$" (and should not have been separated into cases). 3. OP was happy with the "less-than case", and appeared to ask about equality. 4. In the manner of a game of telephone, an edit changed the meaning of the question, so that the task ("prove equality") as stated is incorrect. – 2017-01-10
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    I concur with @AndrewD.Hwang's explanation of the events leading to the current sorry state of the question. Note that "Prove that |a+b+c|<|a|+|b|+|c|" is wrong as well. – 2017-01-10
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    @user405589: I rolled back the version and made some re-wording changes that attempt to capture your thinking. If anything doesn't match your intent, please edit to fix. – 2017-01-10

2 Answers 2

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First two critiques on the work so far:

  1. An inequality "$A \leq B$" means "$A < B$ or $A = B$". Splitting the proof into "cases" is usually not necessary, and is not necessary here.

    If splitting were necessary, the structure of a typical proof would be: "If $A = B$, there is nothing to prove. Otherwise, $A \neq B$. [Proof that if $A \neq B$, then $A < B$.]"

    Note, incidentally, that "$A < B$ and $A = B$" is a contradiction: The usual meanings of the less-than and equality symbols are logically incompatible. This highlights that fact that the proof cannot be separated into first proving strict inequality and then equality.

  2. If $a$, $b$, and $c$ are vectors, there is no reason that $a + b + c = 0$, and therefore no reason that they constitute sides of a triangle.

    If $A$, $B$, and $C$ are points such that $a = OA$, etc., the sides of the triangle are $(B - A)$, $(C - B)$, and $(C - A)$, whose sum is zero. Since these vectors are (for present purposes) unrelated to $a$, $b$, and $c$, the construction in your diagram turns out to be unhelpful.

Instead, you probably have seen the ordinary triangle inequality $$ |a + b| \leq |a| + |b|\quad\text{for all vectors $a$ and $b$.} \tag{1} $$ To prove the triangle inequality for three summands, apply (1) recursively to $(a + b) + c$. Can you take it from there?

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The triangle inequality states that $|a+b|\leq|a|+|b|$. We just apply this twice to get:

$$|(a+b)+c|\leq|a+b|+|c|\leq |a|+|b|+|c|$$ and we're done.