Prove $2^n > 1 + n sqrt2^n-1$ for all $n > 2$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Inequality. $sumsqrtx^2+xy+y^2geq sumsqrt2x^2+xy.$Prove the inequality $sum_k=1^2n-1sqrtk(4n-k)<pi n^2$ for all natural $n$Prove inequality $|sqrtx- sqrty| le sqrtx-y$Two sequences defined by recurrence relations satisfy $x_n/y_n<sqrt7$ for all $n$Prove that $sqrt[n]n>sqrt[n+1]n+1$ for all $n geq 3$Given $a,b,cge1;abcge8$. Proving that $sqrta^2-1+sqrtb^2-1+sqrtc^2-1ge 3sqrt3$For all nonnegative real numbers $x,y$ and $z$, prove that $dfrac(x+y+z)^23 geq xsqrtyz+ysqrtxz+zsqrtxy.$Suppose $0<a<b$. Prove for all $ngeq 2$, $0< sqrt[n]a< sqrt[n]b$.Prove that $ln n < sqrt n$ for all natural numbers n by inductionMonotonicity of function proof: Prove that $e^arctan x(x+sqrt1+x^2) < e^2x$ for all $x >0$
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how can a perfect fourth interval be considered either consonant or dissonant?
Prove $2^n > 1 + n sqrt2^n-1$ for all $n > 2$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Inequality. $sumsqrtx^2+xy+y^2geq sumsqrt2x^2+xy.$Prove the inequality $sum_k=1^2n-1sqrtk(4n-k)<pi n^2$ for all natural $n$Prove inequality $|sqrtx- sqrty| le sqrtx-y$Two sequences defined by recurrence relations satisfy $x_n/y_n<sqrt7$ for all $n$Prove that $sqrt[n]n>sqrt[n+1]n+1$ for all $n geq 3$Given $a,b,cge1;abcge8$. Proving that $sqrta^2-1+sqrtb^2-1+sqrtc^2-1ge 3sqrt3$For all nonnegative real numbers $x,y$ and $z$, prove that $dfrac(x+y+z)^23 geq xsqrtyz+ysqrtxz+zsqrtxy.$Suppose $0<a<b$. Prove for all $ngeq 2$, $0< sqrt[n]a< sqrt[n]b$.Prove that $ln n < sqrt n$ for all natural numbers n by inductionMonotonicity of function proof: Prove that $e^arctan x(x+sqrt1+x^2) < e^2x$ for all $x >0$
$begingroup$
Prove that $$2^n> 1 + n sqrt2^n - 1 text for all n > 2.$$
inequality
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
$endgroup$
add a comment |
$begingroup$
Prove that $$2^n> 1 + n sqrt2^n - 1 text for all n > 2.$$
inequality
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
$endgroup$
$begingroup$
what exactly is your problem with this exercise?
$endgroup$
– Pink Panther
Mar 31 at 14:43
3
$begingroup$
Check it for $n=3$. Then intuitively the left side doubles each time $n$ increases by $1$ while the right is multiplied by $frac n+1nsqrt 2 lt 2$. Can you convert that to an induction proof?
$endgroup$
– Ross Millikan
Mar 31 at 14:52
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– R_B
Mar 31 at 15:03
add a comment |
$begingroup$
Prove that $$2^n> 1 + n sqrt2^n - 1 text for all n > 2.$$
inequality
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
$endgroup$
Prove that $$2^n> 1 + n sqrt2^n - 1 text for all n > 2.$$
inequality
inequality
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
edited Mar 31 at 15:47
Robert Soupe
11.5k21950
11.5k21950
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
asked Mar 31 at 14:42
Shashwat SharmaShashwat Sharma
4
4
$begingroup$
what exactly is your problem with this exercise?
$endgroup$
– Pink Panther
Mar 31 at 14:43
3
$begingroup$
Check it for $n=3$. Then intuitively the left side doubles each time $n$ increases by $1$ while the right is multiplied by $frac n+1nsqrt 2 lt 2$. Can you convert that to an induction proof?
$endgroup$
– Ross Millikan
Mar 31 at 14:52
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– R_B
Mar 31 at 15:03
add a comment |
$begingroup$
what exactly is your problem with this exercise?
$endgroup$
– Pink Panther
Mar 31 at 14:43
3
$begingroup$
Check it for $n=3$. Then intuitively the left side doubles each time $n$ increases by $1$ while the right is multiplied by $frac n+1nsqrt 2 lt 2$. Can you convert that to an induction proof?
$endgroup$
– Ross Millikan
Mar 31 at 14:52
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– R_B
Mar 31 at 15:03
$begingroup$
what exactly is your problem with this exercise?
$endgroup$
– Pink Panther
Mar 31 at 14:43
$begingroup$
what exactly is your problem with this exercise?
$endgroup$
– Pink Panther
Mar 31 at 14:43
3
3
$begingroup$
Check it for $n=3$. Then intuitively the left side doubles each time $n$ increases by $1$ while the right is multiplied by $frac n+1nsqrt 2 lt 2$. Can you convert that to an induction proof?
$endgroup$
– Ross Millikan
Mar 31 at 14:52
$begingroup$
Check it for $n=3$. Then intuitively the left side doubles each time $n$ increases by $1$ while the right is multiplied by $frac n+1nsqrt 2 lt 2$. Can you convert that to an induction proof?
$endgroup$
– Ross Millikan
Mar 31 at 14:52
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– R_B
Mar 31 at 15:03
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– R_B
Mar 31 at 15:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $n=3$, you have $2^3=8$, and $1+3sqrt2 leq 1 + 3 times 1,5 = 5,5$ so the statement is true.
Let's suppose that you have
$2^n > 1 + n sqrt2^n-1$ for an integer $n geq 3$. Then
$$2^n+1 > 2 + 2n sqrt2^n-1 = 2 + sqrt2times nsqrt2^n > 1 + sqrt2times nsqrt2^n$$
And for all $n > 2$, you have $$n > frac1sqrt2-1$$ so $n sqrt2 > n+1$, so you deduce
$$2^n+1 > 1 + (n+1) sqrt2^n$$
and you are done.
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
$endgroup$
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
For $n=3$, you have $2^3=8$, and $1+3sqrt2 leq 1 + 3 times 1,5 = 5,5$ so the statement is true.
Let's suppose that you have
$2^n > 1 + n sqrt2^n-1$ for an integer $n geq 3$. Then
$$2^n+1 > 2 + 2n sqrt2^n-1 = 2 + sqrt2times nsqrt2^n > 1 + sqrt2times nsqrt2^n$$
And for all $n > 2$, you have $$n > frac1sqrt2-1$$ so $n sqrt2 > n+1$, so you deduce
$$2^n+1 > 1 + (n+1) sqrt2^n$$
and you are done.
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
$endgroup$
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
add a comment |
$begingroup$
For $n=3$, you have $2^3=8$, and $1+3sqrt2 leq 1 + 3 times 1,5 = 5,5$ so the statement is true.
Let's suppose that you have
$2^n > 1 + n sqrt2^n-1$ for an integer $n geq 3$. Then
$$2^n+1 > 2 + 2n sqrt2^n-1 = 2 + sqrt2times nsqrt2^n > 1 + sqrt2times nsqrt2^n$$
And for all $n > 2$, you have $$n > frac1sqrt2-1$$ so $n sqrt2 > n+1$, so you deduce
$$2^n+1 > 1 + (n+1) sqrt2^n$$
and you are done.
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
$endgroup$
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
add a comment |
$begingroup$
For $n=3$, you have $2^3=8$, and $1+3sqrt2 leq 1 + 3 times 1,5 = 5,5$ so the statement is true.
Let's suppose that you have
$2^n > 1 + n sqrt2^n-1$ for an integer $n geq 3$. Then
$$2^n+1 > 2 + 2n sqrt2^n-1 = 2 + sqrt2times nsqrt2^n > 1 + sqrt2times nsqrt2^n$$
And for all $n > 2$, you have $$n > frac1sqrt2-1$$ so $n sqrt2 > n+1$, so you deduce
$$2^n+1 > 1 + (n+1) sqrt2^n$$
and you are done.
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
$endgroup$
For $n=3$, you have $2^3=8$, and $1+3sqrt2 leq 1 + 3 times 1,5 = 5,5$ so the statement is true.
Let's suppose that you have
$2^n > 1 + n sqrt2^n-1$ for an integer $n geq 3$. Then
$$2^n+1 > 2 + 2n sqrt2^n-1 = 2 + sqrt2times nsqrt2^n > 1 + sqrt2times nsqrt2^n$$
And for all $n > 2$, you have $$n > frac1sqrt2-1$$ so $n sqrt2 > n+1$, so you deduce
$$2^n+1 > 1 + (n+1) sqrt2^n$$
and you are done.
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
answered Mar 31 at 15:19
TheSilverDoeTheSilverDoe
5,566216
5,566216
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
add a comment |
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
It is simply said to use principles of inequalities it's not specified which prove this problem!
$endgroup$
– Shashwat Sharma
Apr 9 at 13:01
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
$begingroup$
@ShashwatSharma What do you mean ?...
$endgroup$
– TheSilverDoe
Apr 9 at 14:27
add a comment |
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$begingroup$
what exactly is your problem with this exercise?
$endgroup$
– Pink Panther
Mar 31 at 14:43
3
$begingroup$
Check it for $n=3$. Then intuitively the left side doubles each time $n$ increases by $1$ while the right is multiplied by $frac n+1nsqrt 2 lt 2$. Can you convert that to an induction proof?
$endgroup$
– Ross Millikan
Mar 31 at 14:52
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– R_B
Mar 31 at 15:03