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Prove $sqrtx^2+1 + frac1sqrtx^2 +1geq 2$ [duplicate]
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How to prove this inequality $ x + frac1x geq 2 $
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I want to show why the last inequality in the problem below $sqrtx^2+1 + frac1sqrtx^2 +1geq 2$ holds. It's clear that $x^2geq 0$ and that equality holds when $x=0$ but how can I clearly show this. The left hand side, for $x^2>0$ is greater than $1$ but the right hand side becomes less than $1$ as $x^2>0$
inequality proof-writing contest-math
asked Jul 30 at 6:40
john fowles
1,088817
marked as duplicate by Martin R, Isaac Browne, Strants, Batominovski, José Carlos Santos Jul 30 at 22:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
How to prove this inequality $ x + frac1x geq 2 $
17 answers
I want to show why the last inequality in the problem below $sqrtx^2+1 + frac1sqrtx^2 +1geq 2$ holds. It's clear that $x^2geq 0$ and that equality holds when $x=0$ but how can I clearly show this. The left hand side, for $x^2>0$ is greater than $1$ but the right hand side becomes less than $1$ as $x^2>0$
inequality proof-writing contest-math
asked Jul 30 at 6:40
john fowles
1,088817
marked as duplicate by Martin R, Isaac Browne, Strants, Batominovski, José Carlos Santos Jul 30 at 22:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Hint: think AM-GM, or any of several related ways.
– dxiv
Jul 30 at 6:42
I don't see how to use AM-GM here?
– john fowles
Jul 30 at 6:48
1
By AM-GM $;displaystyle frac;a + dfrac1a;2 ge sqrta cdot frac1a = 1,$, then use it for $,a=sqrtx^2+1,$.
– dxiv
Jul 30 at 6:51
This is very nice!
– john fowles
Jul 30 at 6:53
Also: Proving an inequality involving square roots and polynomials $fraca^2+2sqrta^2+1ge2$.
– Martin R
Jul 30 at 8:10
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
How to prove this inequality $ x + frac1x geq 2 $
17 answers
I want to show why the last inequality in the problem below $sqrtx^2+1 + frac1sqrtx^2 +1geq 2$ holds. It's clear that $x^2geq 0$ and that equality holds when $x=0$ but how can I clearly show this. The left hand side, for $x^2>0$ is greater than $1$ but the right hand side becomes less than $1$ as $x^2>0$
inequality proof-writing contest-math
asked Jul 30 at 6:40
john fowles
1,088817
This question already has an answer here:
How to prove this inequality $ x + frac1x geq 2 $
17 answers
I want to show why the last inequality in the problem below $sqrtx^2+1 + frac1sqrtx^2 +1geq 2$ holds. It's clear that $x^2geq 0$ and that equality holds when $x=0$ but how can I clearly show this. The left hand side, for $x^2>0$ is greater than $1$ but the right hand side becomes less than $1$ as $x^2>0$
This question already has an answer here:
How to prove this inequality $ x + frac1x geq 2 $
17 answers
inequality proof-writing contest-math
asked Jul 30 at 6:40
john fowles
1,088817
asked Jul 30 at 6:40
john fowles
1,088817
asked Jul 30 at 6:40
john fowles
1,088817
asked Jul 30 at 6:40
john fowles
1,088817
1,088817
marked as duplicate by Martin R, Isaac Browne, Strants, Batominovski, José Carlos Santos Jul 30 at 22:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Isaac Browne, Strants, Batominovski, José Carlos Santos Jul 30 at 22:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Hint: think AM-GM, or any of several related ways.
– dxiv
Jul 30 at 6:42
I don't see how to use AM-GM here?
– john fowles
Jul 30 at 6:48
1
By AM-GM $;displaystyle frac;a + dfrac1a;2 ge sqrta cdot frac1a = 1,$, then use it for $,a=sqrtx^2+1,$.
– dxiv
Jul 30 at 6:51
This is very nice!
– john fowles
Jul 30 at 6:53
Also: Proving an inequality involving square roots and polynomials $fraca^2+2sqrta^2+1ge2$.
– Martin R
Jul 30 at 8:10
add a comment |Â
Hint: think AM-GM, or any of several related ways.
– dxiv
Jul 30 at 6:42
I don't see how to use AM-GM here?
– john fowles
Jul 30 at 6:48
1
By AM-GM $;displaystyle frac;a + dfrac1a;2 ge sqrta cdot frac1a = 1,$, then use it for $,a=sqrtx^2+1,$.
– dxiv
Jul 30 at 6:51
This is very nice!
– john fowles
Jul 30 at 6:53
Also: Proving an inequality involving square roots and polynomials $fraca^2+2sqrta^2+1ge2$.
– Martin R
Jul 30 at 8:10
Hint: think AM-GM, or any of several related ways.
– dxiv
Jul 30 at 6:42
Hint: think AM-GM, or any of several related ways.
– dxiv
Jul 30 at 6:42
I don't see how to use AM-GM here?
– john fowles
Jul 30 at 6:48
I don't see how to use AM-GM here?
– john fowles
Jul 30 at 6:48
1
1
By AM-GM $;displaystyle frac;a + dfrac1a;2 ge sqrta cdot frac1a = 1,$, then use it for $,a=sqrtx^2+1,$.
– dxiv
Jul 30 at 6:51
By AM-GM $;displaystyle frac;a + dfrac1a;2 ge sqrta cdot frac1a = 1,$, then use it for $,a=sqrtx^2+1,$.
– dxiv
Jul 30 at 6:51
This is very nice!
– john fowles
Jul 30 at 6:53
This is very nice!
– john fowles
Jul 30 at 6:53
Also: Proving an inequality involving square roots and polynomials $fraca^2+2sqrta^2+1ge2$.
– Martin R
Jul 30 at 8:10
Also: Proving an inequality involving square roots and polynomials $fraca^2+2sqrta^2+1ge2$.
– Martin R
Jul 30 at 8:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The inequality $(y-1)^2 geq 0$ gives $y+frac 1 y geq 2$ for any positive number $y$. Take $y=sqrt 1+x^2$.
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
2
down vote
The derivative of $f(y)=y+1/y$ is $1-1/y^2$, so the function $f$ increases in $[1,infty)$, hence $f(y)geq f(1)=2$.
answered Jul 30 at 6:43
uniquesolution
7,526721
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.
– dxiv
Jul 30 at 6:45
1
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
1
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The inequality $(y-1)^2 geq 0$ gives $y+frac 1 y geq 2$ for any positive number $y$. Take $y=sqrt 1+x^2$.
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
3
down vote
accepted
The inequality $(y-1)^2 geq 0$ gives $y+frac 1 y geq 2$ for any positive number $y$. Take $y=sqrt 1+x^2$.
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The inequality $(y-1)^2 geq 0$ gives $y+frac 1 y geq 2$ for any positive number $y$. Take $y=sqrt 1+x^2$.
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
The inequality $(y-1)^2 geq 0$ gives $y+frac 1 y geq 2$ for any positive number $y$. Take $y=sqrt 1+x^2$.
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:43
Kavi Rama Murthy
19.7k2829
19.7k2829
add a comment |Â
add a comment |Â
up vote
2
down vote
The derivative of $f(y)=y+1/y$ is $1-1/y^2$, so the function $f$ increases in $[1,infty)$, hence $f(y)geq f(1)=2$.
answered Jul 30 at 6:43
uniquesolution
7,526721
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.
– dxiv
Jul 30 at 6:45
1
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
1
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
add a comment |Â
up vote
2
down vote
The derivative of $f(y)=y+1/y$ is $1-1/y^2$, so the function $f$ increases in $[1,infty)$, hence $f(y)geq f(1)=2$.
answered Jul 30 at 6:43
uniquesolution
7,526721
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.
– dxiv
Jul 30 at 6:45
1
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
1
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The derivative of $f(y)=y+1/y$ is $1-1/y^2$, so the function $f$ increases in $[1,infty)$, hence $f(y)geq f(1)=2$.
answered Jul 30 at 6:43
uniquesolution
7,526721
The derivative of $f(y)=y+1/y$ is $1-1/y^2$, so the function $f$ increases in $[1,infty)$, hence $f(y)geq f(1)=2$.
answered Jul 30 at 6:43
uniquesolution
7,526721
answered Jul 30 at 6:43
uniquesolution
7,526721
answered Jul 30 at 6:43
uniquesolution
7,526721
answered Jul 30 at 6:43
uniquesolution
7,526721
7,526721
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.
– dxiv
Jul 30 at 6:45
1
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
1
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
add a comment |Â
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.
– dxiv
Jul 30 at 6:45
1
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
1
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.– dxiv
Jul 30 at 6:45
f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument.– dxiv
Jul 30 at 6:45
1
1
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
@dxiv - Since $sqrtx^2+1geq 1$ for all $xinmathbbR$, the interval $[0,1]$ is irrelevant.
– uniquesolution
Jul 30 at 6:51
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
Right, but that belongs in the answer, not a comment.
– dxiv
Jul 30 at 6:53
1
1
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
Feel free to include this in your answer.
– uniquesolution
Jul 30 at 6:53
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
I didn't post an answer. Feel free to spell it out in your answer.
– dxiv
Jul 30 at 6:54
add a comment |Â
Hint: think AM-GM, or any of several related ways.
– dxiv
Jul 30 at 6:42
I don't see how to use AM-GM here?
– john fowles
Jul 30 at 6:48
1
By AM-GM $;displaystyle frac;a + dfrac1a;2 ge sqrta cdot frac1a = 1,$, then use it for $,a=sqrtx^2+1,$.
– dxiv
Jul 30 at 6:51
This is very nice!
– john fowles
Jul 30 at 6:53
Also: Proving an inequality involving square roots and polynomials $fraca^2+2sqrta^2+1ge2$.
– Martin R
Jul 30 at 8:10