Prove that |a+b|≤|a|+|b| where a,b are real numbers
We can start by considering two cases:
Case 1: a+b ≥ 0
In this case, we have
|a+b| = a+b
|a|+|b| = a+b
So it is clear that |a+b| ≤ |a|+|b|.
Case 2: a+b < 0
In this case, we have
|a+b| = -(a+b)
|a|+|b| = a+(-b), since |b| = -b in this case
Adding the two inequalities gives
|a+b| + (|a|+|b|) ≤ a+b + (a-b) = 2a ≤ 2|a|
Therefore, we have
|a+b| ≤ 2|a| - |b|
It remains to show that 2|a| - |b| is nonnegative. Since |a| is nonnegative, we have
2|a| - |b| ≥ 2|a| - |a| = |a| ≥ 0
Thus, we have shown that |a+b| ≤ |a|+|b| in both cases, and the result follows.
To prove that |a+b|≤|a|+|b|, we can consider two cases:
Case 1: a+b ≥ 0
If a+b ≥ 0, then |a+b| = a+b, |a| = a, and |b| = b. In this case, the inequality |a+b|≤|a|+|b| becomes a+b ≤ a+b, which is always true.
Case 2: a+b < 0
If a+b < 0, then |a+b| = -(a+b), |a| = -a, and |b| = -b. In this case, the inequality |a+b|≤|a|+|b| becomes -(a+b) ≤ -a -b. We can simplify this to -a - b ≤ -a - b, which is also always true.
Since the inequality |a+b|≤|a|+|b| is true in both cases, it holds for all real numbers a and b. Therefore, it is proven that |a+b|≤|a|+|b| for any real numbers a and b.