Show that $frac{1}{x+y}leqfrac{1}{x+z}+frac{1}{y+z}+1$
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I have the following problem, if $xneq yneq zneq 0$, with $x,y,zinmathbb{N}$. If we define $f:mathbb{N}timesmathbb{N}tomathbb{R}$ as, $$f(x,y)=begin{cases} qquad 0 qquadqquad, x=y\ 1+dfrac{1}{x+y}qquad , xneq yend{cases}$$ I need prove that $f(x,y)leq f(x,z)+f(y,z)$.
My attemp, Let $x,y,zinmathbb{N}$ differents, then I need prove that $$dfrac{1}{x+y}leqdfrac{1}{x+z}+dfrac{1}{y+z}+1$$
My idea was find some inequalities, that maybe can help me, for instance $dfrac{1}{x}+dfrac{1}{y}geqdfrac{2}{x+y}$, but if I take $x=m+n$ and $y=n+k$, then $$dfrac{1}{m+k}+dfrac{1}{n+k}geqdfrac{2}{m+n+2k}$$
If I prove that $dfrac{2}{m+n+2k}>dfrac{1}{m+n}$ I finished the problem, but this implies that $0>2k$, but $kinmathbb{N}$ So, I don't know how prove this, Any hint will be appreciated. Thanks!
calculus
asked Nov 18 at 2:13
Johnny.c
342
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up vote
1
down vote
favorite
I have the following problem, if $xneq yneq zneq 0$, with $x,y,zinmathbb{N}$. If we define $f:mathbb{N}timesmathbb{N}tomathbb{R}$ as, $$f(x,y)=begin{cases} qquad 0 qquadqquad, x=y\ 1+dfrac{1}{x+y}qquad , xneq yend{cases}$$ I need prove that $f(x,y)leq f(x,z)+f(y,z)$.
My attemp, Let $x,y,zinmathbb{N}$ differents, then I need prove that $$dfrac{1}{x+y}leqdfrac{1}{x+z}+dfrac{1}{y+z}+1$$
My idea was find some inequalities, that maybe can help me, for instance $dfrac{1}{x}+dfrac{1}{y}geqdfrac{2}{x+y}$, but if I take $x=m+n$ and $y=n+k$, then $$dfrac{1}{m+k}+dfrac{1}{n+k}geqdfrac{2}{m+n+2k}$$
If I prove that $dfrac{2}{m+n+2k}>dfrac{1}{m+n}$ I finished the problem, but this implies that $0>2k$, but $kinmathbb{N}$ So, I don't know how prove this, Any hint will be appreciated. Thanks!
calculus
asked Nov 18 at 2:13
Johnny.c
342
1
Unless I'm missing something, the left side is always less than $1$, whereas the right side is greater than $1$.
– Theo Bendit
Nov 18 at 2:23
mm ok, This problem was proposed in class, my teacher define a metric $f(x,y)$ on $mathbb{N}timesmathbb{N}$ where $x,yinmathbb{N}$, so the problem is show that the triangle inequality. Thanks!
– Johnny.c
Nov 18 at 2:28
It's definitely not a metric, since $f(x, x) > 0$. I'm not sure what your teacher was intending.
– Theo Bendit
Nov 18 at 2:29
Sorry, I forgot mentioned the cases when $x= y$.
– Johnny.c
Nov 18 at 2:34
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following problem, if $xneq yneq zneq 0$, with $x,y,zinmathbb{N}$. If we define $f:mathbb{N}timesmathbb{N}tomathbb{R}$ as, $$f(x,y)=begin{cases} qquad 0 qquadqquad, x=y\ 1+dfrac{1}{x+y}qquad , xneq yend{cases}$$ I need prove that $f(x,y)leq f(x,z)+f(y,z)$.
My attemp, Let $x,y,zinmathbb{N}$ differents, then I need prove that $$dfrac{1}{x+y}leqdfrac{1}{x+z}+dfrac{1}{y+z}+1$$
My idea was find some inequalities, that maybe can help me, for instance $dfrac{1}{x}+dfrac{1}{y}geqdfrac{2}{x+y}$, but if I take $x=m+n$ and $y=n+k$, then $$dfrac{1}{m+k}+dfrac{1}{n+k}geqdfrac{2}{m+n+2k}$$
If I prove that $dfrac{2}{m+n+2k}>dfrac{1}{m+n}$ I finished the problem, but this implies that $0>2k$, but $kinmathbb{N}$ So, I don't know how prove this, Any hint will be appreciated. Thanks!
calculus
asked Nov 18 at 2:13
Johnny.c
342
I have the following problem, if $xneq yneq zneq 0$, with $x,y,zinmathbb{N}$. If we define $f:mathbb{N}timesmathbb{N}tomathbb{R}$ as, $$f(x,y)=begin{cases} qquad 0 qquadqquad, x=y\ 1+dfrac{1}{x+y}qquad , xneq yend{cases}$$ I need prove that $f(x,y)leq f(x,z)+f(y,z)$.
My attemp, Let $x,y,zinmathbb{N}$ differents, then I need prove that $$dfrac{1}{x+y}leqdfrac{1}{x+z}+dfrac{1}{y+z}+1$$
My idea was find some inequalities, that maybe can help me, for instance $dfrac{1}{x}+dfrac{1}{y}geqdfrac{2}{x+y}$, but if I take $x=m+n$ and $y=n+k$, then $$dfrac{1}{m+k}+dfrac{1}{n+k}geqdfrac{2}{m+n+2k}$$
If I prove that $dfrac{2}{m+n+2k}>dfrac{1}{m+n}$ I finished the problem, but this implies that $0>2k$, but $kinmathbb{N}$ So, I don't know how prove this, Any hint will be appreciated. Thanks!
calculus
calculus
asked Nov 18 at 2:13
Johnny.c
342
asked Nov 18 at 2:13
Johnny.c
342
edited Nov 18 at 3:00
asked Nov 18 at 2:13
Johnny.c
342
asked Nov 18 at 2:13
Johnny.c
342
asked Nov 18 at 2:13
Johnny.c
342
342
1
Unless I'm missing something, the left side is always less than $1$, whereas the right side is greater than $1$.
– Theo Bendit
Nov 18 at 2:23
mm ok, This problem was proposed in class, my teacher define a metric $f(x,y)$ on $mathbb{N}timesmathbb{N}$ where $x,yinmathbb{N}$, so the problem is show that the triangle inequality. Thanks!
– Johnny.c
Nov 18 at 2:28
It's definitely not a metric, since $f(x, x) > 0$. I'm not sure what your teacher was intending.
– Theo Bendit
Nov 18 at 2:29
Sorry, I forgot mentioned the cases when $x= y$.
– Johnny.c
Nov 18 at 2:34
add a comment |
1
Unless I'm missing something, the left side is always less than $1$, whereas the right side is greater than $1$.
– Theo Bendit
Nov 18 at 2:23
mm ok, This problem was proposed in class, my teacher define a metric $f(x,y)$ on $mathbb{N}timesmathbb{N}$ where $x,yinmathbb{N}$, so the problem is show that the triangle inequality. Thanks!
– Johnny.c
Nov 18 at 2:28
It's definitely not a metric, since $f(x, x) > 0$. I'm not sure what your teacher was intending.
– Theo Bendit
Nov 18 at 2:29
Sorry, I forgot mentioned the cases when $x= y$.
– Johnny.c
Nov 18 at 2:34
1
1
Unless I'm missing something, the left side is always less than $1$, whereas the right side is greater than $1$.
– Theo Bendit
Nov 18 at 2:23
Unless I'm missing something, the left side is always less than $1$, whereas the right side is greater than $1$.
– Theo Bendit
Nov 18 at 2:23
mm ok, This problem was proposed in class, my teacher define a metric $f(x,y)$ on $mathbb{N}timesmathbb{N}$ where $x,yinmathbb{N}$, so the problem is show that the triangle inequality. Thanks!
– Johnny.c
Nov 18 at 2:28
mm ok, This problem was proposed in class, my teacher define a metric $f(x,y)$ on $mathbb{N}timesmathbb{N}$ where $x,yinmathbb{N}$, so the problem is show that the triangle inequality. Thanks!
– Johnny.c
Nov 18 at 2:28
It's definitely not a metric, since $f(x, x) > 0$. I'm not sure what your teacher was intending.
– Theo Bendit
Nov 18 at 2:29
It's definitely not a metric, since $f(x, x) > 0$. I'm not sure what your teacher was intending.
– Theo Bendit
Nov 18 at 2:29
Sorry, I forgot mentioned the cases when $x= y$.
– Johnny.c
Nov 18 at 2:34
Sorry, I forgot mentioned the cases when $x= y$.
– Johnny.c
Nov 18 at 2:34
add a comment |
1 Answer
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Hint: Note that the left hand side of inequality is always less than or equal to 1. So the given inequality follows.
answered Nov 18 at 3:10
Thomas Shelby
675115
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Note that the left hand side of inequality is always less than or equal to 1. So the given inequality follows.
answered Nov 18 at 3:10
Thomas Shelby
675115
add a comment |
up vote
1
down vote
Hint: Note that the left hand side of inequality is always less than or equal to 1. So the given inequality follows.
answered Nov 18 at 3:10
Thomas Shelby
675115
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint: Note that the left hand side of inequality is always less than or equal to 1. So the given inequality follows.
answered Nov 18 at 3:10
Thomas Shelby
675115
Hint: Note that the left hand side of inequality is always less than or equal to 1. So the given inequality follows.
answered Nov 18 at 3:10
Thomas Shelby
675115
answered Nov 18 at 3:10
Thomas Shelby
675115
answered Nov 18 at 3:10
Thomas Shelby
675115
answered Nov 18 at 3:10
Thomas Shelby
675115
675115
add a comment |
add a comment |
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Unless I'm missing something, the left side is always less than $1$, whereas the right side is greater than $1$.
– Theo Bendit
Nov 18 at 2:23
mm ok, This problem was proposed in class, my teacher define a metric $f(x,y)$ on $mathbb{N}timesmathbb{N}$ where $x,yinmathbb{N}$, so the problem is show that the triangle inequality. Thanks!
– Johnny.c
Nov 18 at 2:28
It's definitely not a metric, since $f(x, x) > 0$. I'm not sure what your teacher was intending.
– Theo Bendit
Nov 18 at 2:29
Sorry, I forgot mentioned the cases when $x= y$.
– Johnny.c
Nov 18 at 2:34