let-d-be-an-application-d-R-2-R-d-x-y-ln-1-x-y-1-x-y-shown-that-d-is-a-distance-on-R-2-please-help-especially-on-triangular-inequality-

Question Number 119401 by cantor last updated on 24/Oct/20

\boldsymbol l e t \boldsymbol d \boldsymbol b e \boldsymbol a n \boldsymbol a p p l i c a t i o n

\boldsymbol d : R^{2} \to R_{+}

\boldsymbol d (\boldsymbol x, \boldsymbol y) = \boldsymbol l n (1 + \frac{∣ \boldsymbol x - \boldsymbol y ∣}{1 + ∣ \boldsymbol x - \boldsymbol y ∣})

\boldsymbol s h o w n \boldsymbol t h a t \boldsymbol d \boldsymbol i s \boldsymbol a \boldsymbol d i s t a n c e

\boldsymbol o n R^{2}

\boldsymbol p l e a s e \boldsymbol h e l p

★ \boldsymbol e s p e c i a l l y \boldsymbol o n \boldsymbol t r i a n g u l a r

\boldsymbol i n e q u a l i t y

Answered by mindispower last updated on 24/Oct/20

d (x, y) ⩾ 0, \forall (x, y) \in R^{2}

d (x, y) + d (y, z)

\frac{∣ x - y ∣}{1 + ∣ x - y ∣} = f (∣ x - y ∣)

f (x) = \frac{x}{1 + x}

f (a) + f (b) ⩾ f (a + b), \forall (a . b) \in R_{+}^{2} . . ?

\Leftrightarrow \frac{a}{1 + a} + \frac{b}{1 + b} ⩾ \frac{a + b}{a + b + 1} . . A

\Leftrightarrow \frac{2 a b + a + b}{(1 + a) (1 + b)} ⩾ \frac{a + b}{(a + b + 1)}

\Leftrightarrow \frac{2 a b + a + b}{a + b + 1 + a b} ⩾ \frac{a + b}{a + b + 1}

\frac{a b}{a + b + 1 + a b} + \frac{a b + a + b}{a + b + a b + 1} ⩾ \frac{a + b}{a + b + 1}

f (t) = \frac{t}{t + 1}, f^{'} (t) = \frac{1}{{(1 + t)}^{2}} ⩾ 0 f i n c r e s e

a b + a + b ⩾ a + b, a b ⩾ 0,

\Rightarrow f (a b + a + b) ⩾ f (a + b)

\Leftrightarrow \frac{a b + a + b}{a b + a + b + 1} + \frac{a b}{a b + a + b + 1} ⩾ \frac{a b + a + b}{a b + a + b + 1} ⩾ \frac{a + b}{a + b + 1}

\Rightarrow A i s t r u e

\frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ y - z ∣}{1 + ∣ y - z ∣} ⩾ \frac{∣ x - y ∣ + ∣ y - z ∣}{1 + ∣ x - y ∣ + ∣ y - z ∣} ⩾ \frac{∣ x - z ∣}{1 + ∣ x - z ∣}

s i n c e ∣ x - y ∣ + ∣ y - z ∣⩾∣ x - z ∣

a n d f i s i n c r e a s e f u n c t i o n

l n (1 + a) + l n (1 + b) ⩾ l n (1 + a + b) \dots \dots

t h i s s h o w i n e q u a l i t y

Answered by 1549442205PVT last updated on 24/Oct/20

a) Since 1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣} ⩾ 1 \forall (x, y) \in R^{2},

d (x, y) = \ln (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣}) ⩾ 0 \forall (x, y) \in R^{2}

Furthermore, it is clear d (x, y) = d (y, x)

Hence, for two any points A (x)

and B (y) we define ρ (A, B) = d (x, y)

then ρ (A, B) is the distance between

two points A and B of R^{2} (\boldsymbol q . \boldsymbol e . \boldsymbol d)

b) Suppose A (x), B (y), C (z) are three

arbitrary points which {aren}^{'} t colinear

We will prove that

d (x, y) + d (x, z) > d (y, z) \Leftrightarrow \ln (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣})

+ \ln (1 + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}) > \ln (2 + \frac{∣ y - z ∣}{1 + ∣ y - z ∣})

\Leftrightarrow (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣}) (1 + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}) >

(1 + \frac{∣ y - z ∣}{1 + ∣ y - z ∣}) (1) . Indeed,

(1) \Leftrightarrow 1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}

+ \frac{∣ x - y ∣∣ x - z ∣}{(1 + ∣ x - y ∣) (1 + ∣ x - z ∣)} > 1 + \frac{∣ y - z ∣}{1 + ∣ y - z ∣}

\Leftrightarrow \frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}

\frac{∣ x - y ∣∣ x - z ∣}{(1 + ∣ x - y ∣) (1 + ∣ x - z ∣)} > \frac{∣ y - z ∣}{1 + ∣ y - z ∣} = \frac{1}{1 + \frac{1}{∣ y - z ∣}}

∙ The case ∣ y - z ∣= \min {∣∣ x - y ∣, ∣ x - z ∣, ∣ y - z ∣}

\Rightarrow \frac{∣ x - y ∣}{1 + ∣ x - y ∣} = \frac{1}{1 + \frac{1}{∣ x - y ∣}} ⩾ \frac{1}{1 + \frac{1}{∣ y - z ∣}}

\Rightarrow L . H . S > R . H . S (\boldsymbol q . \boldsymbol e . \boldsymbol d)

∙ The case ∣ y - z ∣= \max {∣∣ x - y ∣, ∣ x - z ∣, ∣ y - z ∣}

\Rightarrow \frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ x - z ∣}{1 + ∣ x - z ∣} ⩾ \frac{∣ x - y ∣ + ∣ x - z ∣}{1 + \max {∣ x - y ∣, ∣ x - z ∣}}

⩾ \frac{∣ y - z ∣}{1 + \max {∣ x - y ∣, ∣ x - z ∣}} ⩾ \frac{∣ y - z ∣}{1 + ∣ y - z ∣}

\Rightarrow L . H . S > R . H . S (\boldsymbol q . \boldsymbol e . \boldsymbol d)

Thus, in every cases we always have

\Rightarrow L . H . S > R . H . S . Therefore, the

trianglar inequality is proved