Question

Prove that |a + b| <= |a| + |b|

Answer 1

This is the triangle inequality.
Its easy to prove by vector algebra.
|a|,|b| and |a+b| are the three sides of a triangle then obviously sum of two sides of a triangle greater than third side.
Algebraically i don't think its hard either. If you ask me it looks kinda obvious.

Answer 2

Often times it's the intuitively obvious things that are the most difficult to rigorously prove :)
Assuming that a and b belong to the real number system, we can assume that the inequality does NOT hold, and then derive a contradiction. Let's therefore assume that
\[ |a + b| > |a| + |b|\]
Both sides are clearly positive, so if we square both sides the inequality still holds, and we find
\[ |a+b|^2 = (a+b)^2 = a^2 + b^2 + 2ab\]
and
\[ (|a| + |b|)^2 = a^2 + b^2 + 2|a||b| \]
so subtracting off the identical terms and dividing by two, our assumption yields
\[ a\cdot b > |a| \cdot |b| \]
We have two cases now. Either a and b have the same sign, or they have different signs. If the signs are the same, then the two sides are equal, and so the statement is false. If they have the same sign, then the left side is the negative of the (clearly positive) right side, and the statement is also false. Therefore, our assumption yields a contradiction. It follows that the negation of our assumption is true, i.e.
\[ |a+b| \leq |a| + |b| \]

Answer 3

Oops. Halfway through that last paragraph it should be
"If they have opposite signs, then the left side is the negative ... "
Not my night with the typos :/

Answer 4

can anyone teach me this in easy words?

Answer 5

As stated above, if you're not looking for rigorous mathematical logic then its relatively easy to convince yourself of the truth of the triangle inequality by imagining that a and b are the legs of a triangle, and a+b is the third leg. It's quite clear that no matter how you arrange a and b, the third leg certainly can't be longer that the sum of their lengths nor can it be shorter than the difference: