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


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

$l e t d b e a n a p p l i c a t i o n$ $d : R^{2} \to R_{+}$ $d (x, y) = l n (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣})$ $s h o w n t h a t d i s a d i s t a n c e$ $o n R^{2}$ $p l e a s e h e l p$ $★ e s p e c i a l l y o n t r i a n g u l a r$ $i n e q u a l i t y$
	Answered by mindispower last updated on 24/Oct/20

	$d (x, y) ⩾ 0, \forall (x, y) \in R^{2}$ $d (x, y) + d (y, z)$ $\frac{∣ x - y ∣}{1 + ∣ x - y ∣} = f (∣ x - y ∣)$ $f (x) = \frac{x}{1 + x}$ $f (a) + f (b) ⩾ f (a + b), \forall (a . b) \in R_{+}^{2} . . ?$ $\Leftrightarrow \frac{a}{1 + a} + \frac{b}{1 + b} ⩾ \frac{a + b}{a + b + 1} . . A$ $\Leftrightarrow \frac{2 a b + a + b}{(1 + a) (1 + b)} ⩾ \frac{a + b}{(a + b + 1)}$ $\Leftrightarrow \frac{2 a b + a + b}{a + b + 1 + a b} ⩾ \frac{a + b}{a + b + 1}$ $\frac{a b}{a + b + 1 + a b} + \frac{a b + a + b}{a + b + a b + 1} ⩾ \frac{a + b}{a + b + 1}$ $f (t) = \frac{t}{t + 1}, f^{'} (t) = \frac{1}{{(1 + t)}^{2}} ⩾ 0 f i n c r e s e$ $a b + a + b ⩾ a + b, a b ⩾ 0,$ $\Rightarrow f (a b + a + b) ⩾ f (a + b)$ $\Leftrightarrow \frac{a b + a + b}{a b + a + b + 1} + \frac{a b}{a b + a + b + 1} ⩾ \frac{a b + a + b}{a b + a + b + 1} ⩾ \frac{a + b}{a + b + 1}$ $\Rightarrow A i s t r u e$ $\frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ y - z ∣}{1 + ∣ y - z ∣} ⩾ \frac{∣ x - y ∣ + ∣ y - z ∣}{1 + ∣ x - y ∣ + ∣ y - z ∣} ⩾ \frac{∣ x - z ∣}{1 + ∣ x - z ∣}$ $s i n c e ∣ x - y ∣ + ∣ y - z ∣⩾∣ x - z ∣$ $a n d f i s i n c r e a s e f u n c t i o n$ $l n (1 + a) + l n (1 + b) ⩾ l n (1 + a + b) . . . . . .$ $t h i s s h o w i n e q u a l i t y$
	Answered by 1549442205PVT last updated on 24/Oct/20
	$a)Since 1+((∣x−y∣)/(1+∣x−y∣))≥1∀(x,y)∈R^2 , d(x,y)=ln(1+((∣x−y∣)/(1+∣x−y∣)))≥0∀(x,y)∈R^2 Furthermore,it is clear d(x,y)=d(y,x) Hence ,for two any points A(x) and B(y)we define ρ(A,B)=d(x,y) then ρ(A,B) is the distance between two points A and B of R^2 (q.e.d) b)Suppose A(x),B(y),C(z) are three arbitrary points which aren′t colinear We will prove that d(x,y)+d(x,z)>d(y,z)⇔ln(1+((∣x−y∣)/(1+∣x−y∣))) +ln(1+((∣x−z∣)/(1+∣x−z∣)))>ln(2+((∣y−z∣)/(1+∣y−z∣))) ⇔(1+((∣x−y∣)/(1+∣x−y∣)))(1+((∣x−z∣)/(1+∣x−z∣)))> (1+((∣y−z∣)/(1+∣y−z∣)))(1).Indeed, (1)⇔1+((∣x−y∣)/(1+∣x−y∣))+((∣x−z∣)/(1+∣x−z∣)) +((∣x−y∣∣x−z∣)/((1+∣x−y∣)(1+∣x−z∣)))>1+((∣y−z∣)/(1+∣y−z∣)) ⇔((∣x−y∣)/(1+∣x−y∣))+((∣x−z∣)/(1+∣x−z∣)) ((∣x−y∣∣x−z∣)/((1+∣x−y∣)(1+∣x−z∣)))>((∣y−z∣)/(1+∣y−z∣))=(1/(1+(1/(∣y−z∣)))) •The case ∣y−z∣=min{∣∣x−y∣,∣x−z∣,∣y−z∣} ⇒((∣x−y∣)/(1+∣x−y∣))=(1/(1+(1/(∣x−y∣))))≥(1/(1+(1/(∣y−z∣)))) ⇒L.H.S>R.H.S(q.e.d) •The case ∣y−z∣=max{∣∣x−y∣,∣x−z∣,∣y−z∣} ⇒((∣x−y∣)/(1+∣x−y∣))+((∣x−z∣)/(1+∣x−z∣))≥((∣x−y∣+∣x−z∣)/(1+max{∣x−y∣,∣x−z∣})) ≥((∣y−z∣)/(1+max{∣x−y∣,∣x−z∣}))≥((∣y−z∣)/(1+∣y−z∣)) ⇒L.H.S>R.H.S(q.e.d) Thus,in every cases we always have ⇒L.H.S>R.H.S.Therefore,the trianglar inequality is proved$
	$a) Since 1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣} ⩾ 1 \forall (x, y) \in R^{2},$ $d (x, y) = \ln (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣}) ⩾ 0 \forall (x, y) \in R^{2}$ $Furthermore, it is clear d (x, y) = d (y, x)$ $Hence, for two any points A (x)$ $and B (y) we define ρ (A, B) = d (x, y)$ $then ρ (A, B) is the distance between$ $two points A and B of R^{2} (q . e . d)$ $b) Suppose A (x), B (y), C (z) are three$ $arbitrary points which {aren}^{'} t colinear$ $We will prove that$ $d (x, y) + d (x, z) > d (y, z) \Leftrightarrow \ln (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣})$ $+ \ln (1 + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}) > \ln (2 + \frac{∣ y - z ∣}{1 + ∣ y - z ∣})$ $\Leftrightarrow (1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣}) (1 + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}) >$ $(1 + \frac{∣ y - z ∣}{1 + ∣ y - z ∣}) (1) . Indeed,$ $(1) \Leftrightarrow 1 + \frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}$ $+ \frac{∣ x - y ∣∣ x - z ∣}{(1 + ∣ x - y ∣) (1 + ∣ x - z ∣)} > 1 + \frac{∣ y - z ∣}{1 + ∣ y - z ∣}$ $\Leftrightarrow \frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ x - z ∣}{1 + ∣ x - z ∣}$ $\frac{∣ x - y ∣∣ x - z ∣}{(1 + ∣ x - y ∣) (1 + ∣ x - z ∣)} > \frac{∣ y - z ∣}{1 + ∣ y - z ∣} = \frac{1}{1 + \frac{1}{∣ y - z ∣}}$ $∙ The case ∣ y - z ∣= \min {∣∣ x - y ∣, ∣ x - z ∣, ∣ y - z ∣}$ $\Rightarrow \frac{∣ x - y ∣}{1 + ∣ x - y ∣} = \frac{1}{1 + \frac{1}{∣ x - y ∣}} ⩾ \frac{1}{1 + \frac{1}{∣ y - z ∣}}$ $\Rightarrow L . H . S > R . H . S (q . e . d)$ $∙ The case ∣ y - z ∣= \max {∣∣ x - y ∣, ∣ x - z ∣, ∣ y - z ∣}$ $\Rightarrow \frac{∣ x - y ∣}{1 + ∣ x - y ∣} + \frac{∣ x - z ∣}{1 + ∣ x - z ∣} ⩾ \frac{∣ x - y ∣ + ∣ x - z ∣}{1 + \max {∣ x - y ∣, ∣ x - z ∣}}$ $⩾ \frac{∣ y - z ∣}{1 + \max {∣ x - y ∣, ∣ x - z ∣}} ⩾ \frac{∣ y - z ∣}{1 + ∣ y - z ∣}$ $\Rightarrow L . H . S > R . H . S (q . e . d)$ $Thus, in every cases we always have$ $\Rightarrow L . H . S > R . H . S . Therefore, the$ $trianglar inequality is proved$
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