Prove using the triangular inequality that: $|a+b| geq |a| - |b|$ The 2019 Stack Overflow Developer Survey Results Are InReverse Triangle Inequality ProofHelp checking proof of reverse triangle inequality $|x| - |y| le |x + y|$?Prove $(a + b)^2 geq 4ab$Prove that $frac1(1+a)^2+frac1(1+b)^2+frac1(1+c)^2+frac1(1+d)^2geq 1$Prove the triangle inequalityProve that this is a metricElementary proof of an inequality with $e^x$ when $|x|<1$.How to prove triangle inequality in How to Prove It Sec. 3.5 Question 12c?Prove the inequality: $1.6^n-2+1.6^n-2 ge 1.6^n-1$Proving that $|a+b| + |a-b| geq |a| + |b|$ for the absolute value functionA confusion about the use of triangular inequality and abolute value in a proofProve that for $a, b in mathbbR$ $|a + b -a| geq |a| - |b-a|$
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Prove using the triangular inequality that: $|a+b| geq |a| - |b|$
The 2019 Stack Overflow Developer Survey Results Are InReverse Triangle Inequality ProofHelp checking proof of reverse triangle inequality $|x| - |y| le |x + y|$?Prove $(a + b)^2 geq 4ab$Prove that $frac1(1+a)^2+frac1(1+b)^2+frac1(1+c)^2+frac1(1+d)^2geq 1$Prove the triangle inequalityProve that this is a metricElementary proof of an inequality with $e^x$ when $|x|<1$.How to prove triangle inequality in How to Prove It Sec. 3.5 Question 12c?Prove the inequality: $1.6^n-2+1.6^n-2 ge 1.6^n-1$Proving that $|a+b| + |a-b| geq |a| + |b|$ for the absolute value functionA confusion about the use of triangular inequality and abolute value in a proofProve that for $a, b in mathbbR$ $|a + b -a| geq |a| - |b-a|$
$begingroup$
How can I prove using the triangular inequality that:
$$|a+b| geq |a| - |b|$$
I already proved it by considering all 8 possible scenarios (like a>b and b=0 ... etc) However I couldn’t manage to find the way to prove it only by using the triangular inequality.
algebra-precalculus absolute-value
asked Mar 23 at 0:16
Facu50196Facu50196
396
$endgroup$
add a comment |
$begingroup$
How can I prove using the triangular inequality that:
$$|a+b| geq |a| - |b|$$
I already proved it by considering all 8 possible scenarios (like a>b and b=0 ... etc) However I couldn’t manage to find the way to prove it only by using the triangular inequality.
algebra-precalculus absolute-value
asked Mar 23 at 0:16
Facu50196Facu50196
396
$endgroup$
2
$begingroup$
This is the "reverse triangle inequality". See here or here for example.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:21
add a comment |
$begingroup$
How can I prove using the triangular inequality that:
$$|a+b| geq |a| - |b|$$
I already proved it by considering all 8 possible scenarios (like a>b and b=0 ... etc) However I couldn’t manage to find the way to prove it only by using the triangular inequality.
algebra-precalculus absolute-value
asked Mar 23 at 0:16
Facu50196Facu50196
396
$endgroup$
How can I prove using the triangular inequality that:
$$|a+b| geq |a| - |b|$$
I already proved it by considering all 8 possible scenarios (like a>b and b=0 ... etc) However I couldn’t manage to find the way to prove it only by using the triangular inequality.
algebra-precalculus absolute-value
algebra-precalculus absolute-value
asked Mar 23 at 0:16
Facu50196Facu50196
396
asked Mar 23 at 0:16
Facu50196Facu50196
396
asked Mar 23 at 0:16
Facu50196Facu50196
396
asked Mar 23 at 0:16
Facu50196Facu50196
396
asked Mar 23 at 0:16
Facu50196Facu50196
396
396
2
$begingroup$
This is the "reverse triangle inequality". See here or here for example.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:21
add a comment |
2
$begingroup$
This is the "reverse triangle inequality". See here or here for example.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:21
2
2
$begingroup$
This is the "reverse triangle inequality". See here or here for example.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:21
$begingroup$
This is the "reverse triangle inequality". See here or here for example.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: the inequality you wish to prove is equivalent to
$$|a + b| + |-b| ge |a|.$$
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
$endgroup$
add a comment |
$begingroup$
By triangular inequality we have:
$$|a|=|a+b +(-b)|leq |a+b|+|b|.$$
Thus,
$$|a+b| geq |a| - |b|.$$
answered Mar 23 at 0:21
S. MathsS. Maths
667116
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: the inequality you wish to prove is equivalent to
$$|a + b| + |-b| ge |a|.$$
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
$endgroup$
add a comment |
$begingroup$
Hint: the inequality you wish to prove is equivalent to
$$|a + b| + |-b| ge |a|.$$
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
$endgroup$
add a comment |
$begingroup$
Hint: the inequality you wish to prove is equivalent to
$$|a + b| + |-b| ge |a|.$$
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
$endgroup$
Hint: the inequality you wish to prove is equivalent to
$$|a + b| + |-b| ge |a|.$$
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
answered Mar 23 at 0:19
Theo BenditTheo Bendit
20.8k12354
20.8k12354
add a comment |
add a comment |
$begingroup$
By triangular inequality we have:
$$|a|=|a+b +(-b)|leq |a+b|+|b|.$$
Thus,
$$|a+b| geq |a| - |b|.$$
answered Mar 23 at 0:21
S. MathsS. Maths
667116
$endgroup$
add a comment |
$begingroup$
By triangular inequality we have:
$$|a|=|a+b +(-b)|leq |a+b|+|b|.$$
Thus,
$$|a+b| geq |a| - |b|.$$
answered Mar 23 at 0:21
S. MathsS. Maths
667116
$endgroup$
add a comment |
$begingroup$
By triangular inequality we have:
$$|a|=|a+b +(-b)|leq |a+b|+|b|.$$
Thus,
$$|a+b| geq |a| - |b|.$$
answered Mar 23 at 0:21
S. MathsS. Maths
667116
$endgroup$
By triangular inequality we have:
$$|a|=|a+b +(-b)|leq |a+b|+|b|.$$
Thus,
$$|a+b| geq |a| - |b|.$$
answered Mar 23 at 0:21
S. MathsS. Maths
667116
answered Mar 23 at 0:21
S. MathsS. Maths
667116
answered Mar 23 at 0:21
S. MathsS. Maths
667116
answered Mar 23 at 0:21
S. MathsS. Maths
667116
667116
add a comment |
add a comment |
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$begingroup$
This is the "reverse triangle inequality". See here or here for example.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:21