# Triangle inequality

In geometry , the triangle inequality is a theorem that says that one side of a triangle is at most as long as the sum of the other two sides. The “at most” includes the special case of equality. The triangle inequality also plays an important role in other areas of mathematics such as linear algebra or functional analysis .

## Shapes of the triangle inequality

### Triangle inequality for triangles

After the triangle is in the triangle , the sum of the lengths of two sides and always at least as great as the length of the third side . This means formally: ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle c}$

${\ displaystyle c \ leq a + b}$

One can also say that the distance from A to B is always at most as large as the distance from A to C and from C to B combined, or to put it in a popular way: "The direct route is always the shortest."

The equals sign only applies if and sections are from - one also speaks of the triangle being "degenerate". ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle c}$

Since, for reasons of symmetry, the following also applies, it follows , analogously, that the total is ${\ displaystyle a \ leq c + b}$${\ displaystyle from \ leq c}$${\ displaystyle ba \ leq c}$

${\ displaystyle \ left | ab \ right | \ leq c \ leq a + b}$.

The left inequality is also sometimes referred to as the reverse triangle inequality . ${\ displaystyle \ left | ab \ right | \ leq c}$

The triangle inequality characterizes distance and magnitude functions . It is therefore used as an axiom of the abstract distance function in metric spaces .

### Triangle inequality for real numbers

For real numbers the following applies:${\ displaystyle | a + b | \ leq | a | + | b |.}$

proof

Because both sides of the inequality are not negative, squaring is an equivalence transformation :
${\ displaystyle a ^ {2} {+} 2ab {+} b ^ {2} \ \ leq \ a ^ {2} {+} 2 {| ab |} {+} b ^ {2}.}$
By deleting identical terms, we arrive at the equivalent inequality
${\ displaystyle 2ab \ leq 2 | ab |.}$
This inequality holds because for any${\ displaystyle x \ leq {| x |}}$${\ displaystyle x \ in \ mathbb {R}.}$

#### Inverse triangle inequality

As with the triangle, an inverse triangle inequality can be derived:

There is a substitution of there ${\ displaystyle | a + b | - | b | \ leq | a |.}$${\ displaystyle a: = x + y, \ b: = - y}$

${\ displaystyle | x | - | y | \ leq | x + y |,}$

if you set it instead, it results ${\ displaystyle b: = - x}$

${\ displaystyle | y | - | x | \ leq | x + y |,}$

so together (because for any real numbers and with and also applies ) ${\ displaystyle u}$${\ displaystyle c}$${\ displaystyle u \ leq c}$${\ displaystyle -u \ leq c}$${\ displaystyle | u | \ leq c}$

${\ displaystyle {\ Big |} | x | - | y | {\ Big |} \ leq | x + y | \ leq | x | + | y ​​|.}$

If you replace with so you also get ${\ displaystyle y}$${\ displaystyle -y,}$

${\ displaystyle {\ Big |} | x | - | y | {\ Big |} \ leq | xy | \ leq | x | + | y ​​|.}$

So overall

${\ displaystyle {\ Big |} | x | - | y | {\ Big |} \ leq | x \ pm y | \ leq | x | + | y ​​|}$ for all ${\ displaystyle x, \, y \ in \ mathbb {R}.}$

### Triangle inequality for complex numbers

The following applies to complex numbers :

${\ displaystyle | z_ {1} {} + z_ {2} | \ leq | z_ {1} | {+} | z_ {2} |.}$

proof

Since all sides are nonnegative, squaring is an equivalence transformation and one obtains
${\ displaystyle z_ {1} {\ overline {z_ {1}}} {+} z_ {1} {\ overline {z_ {2}}} {+} {\ underbrace {{\ overline {z_ {1}} } z_ {2}} _ {= {\ overline {z_ {1} {\ overline {z_ {2}}}}}}} {+} z_ {2} {\ overline {z_ {2}}} \ \ leq \ z_ {1} {\ overline {z_ {1}}} {+} 2 {\ underbrace {| z_ {1} z_ {2} |} _ {= | z_ {1} {\ overline {z_ {2 }}} |}} {+} z_ {2} {\ overline {z_ {2}}},}$
where the overline means complex conjugation . If you delete identical terms and set so remains ${\ displaystyle z {\ mathrel {: = \,}} z_ {1} {\ overline {z_ {2}}},}$
${\ displaystyle z {+} {\ bar {z}} \ leq 2 {| z |}}$
to show. With you get ${\ displaystyle z = u {+} iv}$
${\ displaystyle (u {+} iv) {+} (u {-} iv) = 2u \ leq 2 {\ sqrt {u ^ {2} {+} v ^ {2}}}}$
or.
${\ displaystyle | u | \ leq {\ sqrt {u ^ {2} {+} v ^ {2}}},}$
which is always fulfilled because of and the monotony of the (real) root function.${\ displaystyle 0 \ leq v ^ {2} \}$

As in the real case, this inequality also follows

${\ displaystyle {\ Big |} | z_ {1} | {-} | z_ {2} | {\ Big |} \ leq | z_ {1} {\ pm} z_ {2} | \ leq | z_ {1 } | {+} | z_ {2} |}$ for all ${\ displaystyle z_ {1}, \, z_ {2} \ in \ mathbb {C}.}$

### Triangle inequality of absolute value functions for bodies

Along with other demands, an amount function for a body is also given by the ${\ displaystyle K}$

 Triangle inequality ${\ displaystyle \ varphi (x + y) \ leq \ varphi (x) + \ varphi (y)}$

established. It has to apply to all If all requirements (see article amount function ) are fulfilled, then an amount function is for the body${\ displaystyle x, y \ in K.}$${\ displaystyle \ varphi}$${\ displaystyle K.}$

Is for all whole , then the amount is called non-Archimedean , otherwise Archimedean . ${\ displaystyle \ varphi (n) \ leq 1}$${\ displaystyle n: = \ underbrace {1+ \ dots +1} _ {n {\ text {times}}}}$

For non-Archimedean amounts, the

 tightened triangle inequality ${\ displaystyle \ varphi (x + y) \ leq \ max (\ varphi (x), \ varphi (y)).}$

It makes the amount an ultrametric . Conversely, any ultrametric amount is non-Archimedean.

### Triangle inequality for sums and integrals

Repeated application of the triangle inequality or complete induction results

${\ displaystyle \ left | \ sum _ {i = 1} ^ {n} x_ {i} \ right | \ leq \ sum _ {i = 1} ^ {n} \ left | x_ {i} \ right |}$

for real or complex numbers . This inequality also applies when integrals are considered instead of sums: ${\ displaystyle x_ {i} \;}$

If , where is an interval, Riemann integrable , then it holds ${\ displaystyle f \ colon I \ to \ mathbb {R}}$${\ displaystyle I = [a, b] \,}$

${\ displaystyle \ left | \ int _ {I} f (x) \, dx \ right | \ leq \ int _ {I} | f (x) | \, dx}$.

This also applies to complex-valued functions , cf. Then there is a complex number such that ${\ displaystyle f \ colon I \ to \ mathbb {C}}$${\ displaystyle \ alpha \;}$

${\ displaystyle \ alpha \ int _ {I} f (x) \, dx = \ left | \ int _ {I} f (x) \, dx \ right |}$and .${\ displaystyle | \ alpha | = 1 \;}$

There

${\ displaystyle \ left | \ int _ {I} f (x) \, dx \ right | = \ alpha \ int _ {I} f (x) \, dx = \ int _ {I} \ alpha \, f (x) \, dx = \ int _ {I} \ operatorname {Re} (\ alpha f (x)) \, dx + i \, \ int _ {I} \ operatorname {Im} (\ alpha f (x )) \, dx}$

is real, must be zero. Also applies ${\ displaystyle \ int _ {I} \ operatorname {Im} (\ alpha f (x)) \, dx}$

${\ displaystyle \ operatorname {Re} (\ alpha f (x)) \ leq | \ alpha f (x) | = | f (x) |}$,

so overall

${\ displaystyle \ left | \ int _ {I} f (x) \, dx \ right | = \ int _ {I} \ operatorname {Re} (\ alpha f (x)) \, dx \ leq \ int _ {I} | f (x) | \, dx}$.

### Triangle inequality for vectors

The following applies to vectors :

${\ displaystyle \ left | {\ vec {a}} + {\ vec {b}} \ right | \ leq \ left | {\ vec {a}} \ right | + \ left | {\ vec {b}} \ right |}$.

The validity of this relationship can be seen by squaring it

${\ displaystyle \ left | {\ vec {a}} + {\ vec {b}} \ right | ^ {2} = \ left \ langle {\ vec {a}} + {\ vec {b}}, { \ vec {a}} + {\ vec {b}} \ right \ rangle = \ left | {\ vec {a}} \ right | ^ {2} +2 \ left \ langle {\ vec {a}}, {\ vec {b}} \ right \ rangle + \ left | {\ vec {b}} \ right | ^ {2} \ leq \ left | {\ vec {a}} \ right | ^ {2} +2 \ left | {\ vec {a}} \ right | \ left | {\ vec {b}} \ right | + \ left | {\ vec {b}} \ right | ^ {2} = \ left (\ left | {\ vec {a}} \ right | + \ left | {\ vec {b}} \ right | \ right) ^ {2}}$,

using the Cauchy-Schwarz inequality :

${\ displaystyle \ langle {\ vec {a}}, {\ vec {b}} \ rangle \ leq \ left | {\ vec {a}} \ right | \ cdot \ left | {\ vec {b}} \ right |}$.

Here, too, it follows as in the real case

${\ displaystyle {\ Big |} \ left | {\ vec {a}} \ right | - \ left | {\ vec {b}} \ right | \, \, {\ Big |} \ leq \ left | { \ vec {a}} \ pm {\ vec {b}} \ right | \ leq \ left | {\ vec {a}} \ right | + \ left | {\ vec {b}} \ right |}$

such as

${\ displaystyle \ left | \ sum _ {i = 1} ^ {n} {\ vec {a_ {i}}} \ right | \ leq \ sum _ {i = 1} ^ {n} \ left | {\ vec {a_ {i}}} \ right |.}$

### Triangle inequality for spherical triangles

Two spherical triangles

The triangle inequality generally does not hold in spherical triangles .

However, it applies if you limit yourself to Eulerian triangles, i.e. those in which each side is shorter than half a great circle.

In the adjacent figure, the following applies

${\ displaystyle \ left | ab \ right | \ leq c_ {1} \ leq a + b,}$

however is . ${\ displaystyle c_ {2}> a + b}$

### Triangle inequality for normalized spaces

In a normalized space , the triangle inequality becomes in the form ${\ displaystyle \ left (X, \ | {\ cdot} \ | \ right)}$

${\ displaystyle \ | x + y \ | \ leq \ | x \ | + \ | y \ |}$

required as one of the properties that the standard must meet for everyone . In particular, it also follows here ${\ displaystyle x, y \ in X \;}$

${\ displaystyle {\ Big |} \ | x \ | - \ | y \ | {\ Big |} \ leq \ | x \ pm y \ | \ leq \ | x \ | + \ | y \ |}$

such as

${\ displaystyle \ left \ | \ sum _ {i = 1} ^ {n} x_ {i} \ right \ | \ leq \ sum _ {i = 1} ^ {n} \ | x_ {i} \ |}$for everyone .${\ displaystyle x_ {i} \ in X \;}$

In the special case of L p -spaces , the triangle inequality is called the Minkowski inequality and is proven by means of Hölder's inequality .

### Triangle inequality for metric spaces

In a metric space , an axiom for the abstract distance function is required that the triangle inequality in the form ${\ displaystyle \ left (X, d \ right)}$

${\ displaystyle d (x, y) \ leq d (x, z) + d (z, y)}$

is fulfilled for all . In every metric space, the triangle inequality applies by definition. From this it can be deduced that in a metric space also the reverse triangle inequality ${\ displaystyle x, y, z \ in X}$

${\ displaystyle \ left | d (x, z) -d (z, y) \ right | \ leq d (x, y)}$

applies to all . In addition, the inequality holds for any${\ displaystyle x, y, z \ in X}$${\ displaystyle x_ {i} \ in X \;}$

${\ displaystyle d (x_ {0}, x_ {n}) \ leq \ sum _ {i = 1} ^ {n} d (x_ {i-1}, x_ {i})}$.