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Show the following inequality
$$
sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3.
$$
I tried to rise to power 3 the expresion above, but things went wrong.
algebra-precalculus
asked 1 hour ago
stefano
373
add a comment |Â
up vote
2
down vote
favorite
Show the following inequality
$$
sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3.
$$
I tried to rise to power 3 the expresion above, but things went wrong.
algebra-precalculus
asked 1 hour ago
stefano
373
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Show the following inequality
$$
sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3.
$$
I tried to rise to power 3 the expresion above, but things went wrong.
algebra-precalculus
asked 1 hour ago
stefano
373
Show the following inequality
$$
sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3.
$$
I tried to rise to power 3 the expresion above, but things went wrong.
algebra-precalculus
algebra-precalculus
asked 1 hour ago
stefano
373
asked 1 hour ago
stefano
373
asked 1 hour ago
stefano
373
asked 1 hour ago
stefano
373
asked 1 hour ago
stefano
373
373
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
The function $f(x)=x^1/3$ is strictly concave on $(0,infty)$, that
is
$$(1-t)f(a)+tf(b)<f((1-t)a+tb)$$
if $0<a<b$ and $0<t<1$. In particular
$$fracf(a)2+fracf(b)2<f(a/2+b/2).$$
Now take $a=3-sqrt[3]3$ and $b=3+sqrt[3]3$.
To show that $f(x)=x^1/3$ is strictly concave, observe that
$f''(x)<0$ for all $x>0$, so that $f'(x)$ is strictly decreasing on $(0,infty)$.
By the Mean Value Theorem
beginalign
&f((1-t)a+tb)-(1-t)f(a)-tf(b)\&=
(1-t)[f((1-t)a+tb)-f(a)]-t[f(b)-f((1-t)a+tb)]\
&=(b-a)f'(c)-(b-a)f'(d)>0
endalign
for some $cin(a,(1-t)+tb)$ and some $din((1-t)a+tb,b)$.
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
add a comment |Â
up vote
0
down vote
Consider the function $f(x) = sqrt[3]x$
The gradient of the gradient is constantly decreasing, can be observed on a graph.
Looking at 3, and a small number a,
$f(3+a)-f(3) < f(3) - f(3-a)$
So,
$f(3+a)+f(3-a) < 2f(3) $
Substituting $sqrt[3]3$ into a. We get:
$$sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3$$
If we think of the function as the distance traveled by a car from the origin wrt time, we can see that the gradient (velocity) is constantly reducing. The car never comes to a stop but is slowing down forever. As we slow down, we can see that the distance traveled in the last $sqrt[3]3$ seconds ($f(3) - f(3-a)$) is more than the distance($f(3+a)-f(3)$) traveled in the next $sqrt[3]3$ seconds.
answered 1 hour ago
Neo
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
COMMENT.-It is also true that for all integers $n,m$ greater than $1$ (for $n=1$ there is equality) we have
$$sqrt[n]m+sqrt[n]m+sqrt[n]m-sqrt[n]mlt2sqrt[n]m$$ Try to prove this.
answered 31 mins ago
Piquito
17.5k31335
add a comment |Â
up vote
0
down vote
Another way to show this is to divide both sides by $sqrt[3]3$. Then the left is $sqrt[3]1+a +sqrt[3]1-a$ (for an $a$ which is in $[0,1)$) and we want to show that this is strictly less than $2$.
Let $f(x) = sqrt[3]1+x +sqrt[3]1-x$. Then $f(0)=2$. And it can be easily shown that $f'(x) < 0$ for all $xin (0,1)$ (since $(1+x)^frac23>(1-x)^frac23$ for all $x$ on that interval). Hence $f$ is strictly decreasing on $(0, 1)$, and therefore strictly less than 2.
answered 17 mins ago
bloomers
7261210
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The function $f(x)=x^1/3$ is strictly concave on $(0,infty)$, that
is
$$(1-t)f(a)+tf(b)<f((1-t)a+tb)$$
if $0<a<b$ and $0<t<1$. In particular
$$fracf(a)2+fracf(b)2<f(a/2+b/2).$$
Now take $a=3-sqrt[3]3$ and $b=3+sqrt[3]3$.
To show that $f(x)=x^1/3$ is strictly concave, observe that
$f''(x)<0$ for all $x>0$, so that $f'(x)$ is strictly decreasing on $(0,infty)$.
By the Mean Value Theorem
beginalign
&f((1-t)a+tb)-(1-t)f(a)-tf(b)\&=
(1-t)[f((1-t)a+tb)-f(a)]-t[f(b)-f((1-t)a+tb)]\
&=(b-a)f'(c)-(b-a)f'(d)>0
endalign
for some $cin(a,(1-t)+tb)$ and some $din((1-t)a+tb,b)$.
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
add a comment |Â
up vote
2
down vote
The function $f(x)=x^1/3$ is strictly concave on $(0,infty)$, that
is
$$(1-t)f(a)+tf(b)<f((1-t)a+tb)$$
if $0<a<b$ and $0<t<1$. In particular
$$fracf(a)2+fracf(b)2<f(a/2+b/2).$$
Now take $a=3-sqrt[3]3$ and $b=3+sqrt[3]3$.
To show that $f(x)=x^1/3$ is strictly concave, observe that
$f''(x)<0$ for all $x>0$, so that $f'(x)$ is strictly decreasing on $(0,infty)$.
By the Mean Value Theorem
beginalign
&f((1-t)a+tb)-(1-t)f(a)-tf(b)\&=
(1-t)[f((1-t)a+tb)-f(a)]-t[f(b)-f((1-t)a+tb)]\
&=(b-a)f'(c)-(b-a)f'(d)>0
endalign
for some $cin(a,(1-t)+tb)$ and some $din((1-t)a+tb,b)$.
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The function $f(x)=x^1/3$ is strictly concave on $(0,infty)$, that
is
$$(1-t)f(a)+tf(b)<f((1-t)a+tb)$$
if $0<a<b$ and $0<t<1$. In particular
$$fracf(a)2+fracf(b)2<f(a/2+b/2).$$
Now take $a=3-sqrt[3]3$ and $b=3+sqrt[3]3$.
To show that $f(x)=x^1/3$ is strictly concave, observe that
$f''(x)<0$ for all $x>0$, so that $f'(x)$ is strictly decreasing on $(0,infty)$.
By the Mean Value Theorem
beginalign
&f((1-t)a+tb)-(1-t)f(a)-tf(b)\&=
(1-t)[f((1-t)a+tb)-f(a)]-t[f(b)-f((1-t)a+tb)]\
&=(b-a)f'(c)-(b-a)f'(d)>0
endalign
for some $cin(a,(1-t)+tb)$ and some $din((1-t)a+tb,b)$.
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
The function $f(x)=x^1/3$ is strictly concave on $(0,infty)$, that
is
$$(1-t)f(a)+tf(b)<f((1-t)a+tb)$$
if $0<a<b$ and $0<t<1$. In particular
$$fracf(a)2+fracf(b)2<f(a/2+b/2).$$
Now take $a=3-sqrt[3]3$ and $b=3+sqrt[3]3$.
To show that $f(x)=x^1/3$ is strictly concave, observe that
$f''(x)<0$ for all $x>0$, so that $f'(x)$ is strictly decreasing on $(0,infty)$.
By the Mean Value Theorem
beginalign
&f((1-t)a+tb)-(1-t)f(a)-tf(b)\&=
(1-t)[f((1-t)a+tb)-f(a)]-t[f(b)-f((1-t)a+tb)]\
&=(b-a)f'(c)-(b-a)f'(d)>0
endalign
for some $cin(a,(1-t)+tb)$ and some $din((1-t)a+tb,b)$.
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
edited 54 mins ago
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
answered 1 hour ago
Lord Shark the Unknown
91.4k955118
91.4k955118
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
add a comment |Â
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
Should the second tf(b) just be f(b)?
– bloomers
55 mins ago
add a comment |Â
up vote
0
down vote
Consider the function $f(x) = sqrt[3]x$
The gradient of the gradient is constantly decreasing, can be observed on a graph.
Looking at 3, and a small number a,
$f(3+a)-f(3) < f(3) - f(3-a)$
So,
$f(3+a)+f(3-a) < 2f(3) $
Substituting $sqrt[3]3$ into a. We get:
$$sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3$$
If we think of the function as the distance traveled by a car from the origin wrt time, we can see that the gradient (velocity) is constantly reducing. The car never comes to a stop but is slowing down forever. As we slow down, we can see that the distance traveled in the last $sqrt[3]3$ seconds ($f(3) - f(3-a)$) is more than the distance($f(3+a)-f(3)$) traveled in the next $sqrt[3]3$ seconds.
answered 1 hour ago
Neo
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
Consider the function $f(x) = sqrt[3]x$
The gradient of the gradient is constantly decreasing, can be observed on a graph.
Looking at 3, and a small number a,
$f(3+a)-f(3) < f(3) - f(3-a)$
So,
$f(3+a)+f(3-a) < 2f(3) $
Substituting $sqrt[3]3$ into a. We get:
$$sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3$$
If we think of the function as the distance traveled by a car from the origin wrt time, we can see that the gradient (velocity) is constantly reducing. The car never comes to a stop but is slowing down forever. As we slow down, we can see that the distance traveled in the last $sqrt[3]3$ seconds ($f(3) - f(3-a)$) is more than the distance($f(3+a)-f(3)$) traveled in the next $sqrt[3]3$ seconds.
answered 1 hour ago
Neo
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider the function $f(x) = sqrt[3]x$
The gradient of the gradient is constantly decreasing, can be observed on a graph.
Looking at 3, and a small number a,
$f(3+a)-f(3) < f(3) - f(3-a)$
So,
$f(3+a)+f(3-a) < 2f(3) $
Substituting $sqrt[3]3$ into a. We get:
$$sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3$$
If we think of the function as the distance traveled by a car from the origin wrt time, we can see that the gradient (velocity) is constantly reducing. The car never comes to a stop but is slowing down forever. As we slow down, we can see that the distance traveled in the last $sqrt[3]3$ seconds ($f(3) - f(3-a)$) is more than the distance($f(3+a)-f(3)$) traveled in the next $sqrt[3]3$ seconds.
answered 1 hour ago
Neo
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the function $f(x) = sqrt[3]x$
The gradient of the gradient is constantly decreasing, can be observed on a graph.
Looking at 3, and a small number a,
$f(3+a)-f(3) < f(3) - f(3-a)$
So,
$f(3+a)+f(3-a) < 2f(3) $
Substituting $sqrt[3]3$ into a. We get:
$$sqrt[3]3+sqrt[3]3+sqrt[3]3-sqrt[3]3<2sqrt[3]3$$
If we think of the function as the distance traveled by a car from the origin wrt time, we can see that the gradient (velocity) is constantly reducing. The car never comes to a stop but is slowing down forever. As we slow down, we can see that the distance traveled in the last $sqrt[3]3$ seconds ($f(3) - f(3-a)$) is more than the distance($f(3+a)-f(3)$) traveled in the next $sqrt[3]3$ seconds.
answered 1 hour ago
Neo
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 55 mins ago
answered 1 hour ago
Neo
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Neo
1397
answered 1 hour ago
Neo
1397
1397
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Neo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
0
down vote
COMMENT.-It is also true that for all integers $n,m$ greater than $1$ (for $n=1$ there is equality) we have
$$sqrt[n]m+sqrt[n]m+sqrt[n]m-sqrt[n]mlt2sqrt[n]m$$ Try to prove this.
answered 31 mins ago
Piquito
17.5k31335
add a comment |Â
up vote
0
down vote
COMMENT.-It is also true that for all integers $n,m$ greater than $1$ (for $n=1$ there is equality) we have
$$sqrt[n]m+sqrt[n]m+sqrt[n]m-sqrt[n]mlt2sqrt[n]m$$ Try to prove this.
answered 31 mins ago
Piquito
17.5k31335
add a comment |Â
up vote
0
down vote
up vote
0
down vote
COMMENT.-It is also true that for all integers $n,m$ greater than $1$ (for $n=1$ there is equality) we have
$$sqrt[n]m+sqrt[n]m+sqrt[n]m-sqrt[n]mlt2sqrt[n]m$$ Try to prove this.
answered 31 mins ago
Piquito
17.5k31335
COMMENT.-It is also true that for all integers $n,m$ greater than $1$ (for $n=1$ there is equality) we have
$$sqrt[n]m+sqrt[n]m+sqrt[n]m-sqrt[n]mlt2sqrt[n]m$$ Try to prove this.
answered 31 mins ago
Piquito
17.5k31335
answered 31 mins ago
Piquito
17.5k31335
answered 31 mins ago
Piquito
17.5k31335
answered 31 mins ago
Piquito
17.5k31335
17.5k31335
add a comment |Â
add a comment |Â
up vote
0
down vote
Another way to show this is to divide both sides by $sqrt[3]3$. Then the left is $sqrt[3]1+a +sqrt[3]1-a$ (for an $a$ which is in $[0,1)$) and we want to show that this is strictly less than $2$.
Let $f(x) = sqrt[3]1+x +sqrt[3]1-x$. Then $f(0)=2$. And it can be easily shown that $f'(x) < 0$ for all $xin (0,1)$ (since $(1+x)^frac23>(1-x)^frac23$ for all $x$ on that interval). Hence $f$ is strictly decreasing on $(0, 1)$, and therefore strictly less than 2.
answered 17 mins ago
bloomers
7261210
add a comment |Â
up vote
0
down vote
Another way to show this is to divide both sides by $sqrt[3]3$. Then the left is $sqrt[3]1+a +sqrt[3]1-a$ (for an $a$ which is in $[0,1)$) and we want to show that this is strictly less than $2$.
Let $f(x) = sqrt[3]1+x +sqrt[3]1-x$. Then $f(0)=2$. And it can be easily shown that $f'(x) < 0$ for all $xin (0,1)$ (since $(1+x)^frac23>(1-x)^frac23$ for all $x$ on that interval). Hence $f$ is strictly decreasing on $(0, 1)$, and therefore strictly less than 2.
answered 17 mins ago
bloomers
7261210
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Another way to show this is to divide both sides by $sqrt[3]3$. Then the left is $sqrt[3]1+a +sqrt[3]1-a$ (for an $a$ which is in $[0,1)$) and we want to show that this is strictly less than $2$.
Let $f(x) = sqrt[3]1+x +sqrt[3]1-x$. Then $f(0)=2$. And it can be easily shown that $f'(x) < 0$ for all $xin (0,1)$ (since $(1+x)^frac23>(1-x)^frac23$ for all $x$ on that interval). Hence $f$ is strictly decreasing on $(0, 1)$, and therefore strictly less than 2.
answered 17 mins ago
bloomers
7261210
Another way to show this is to divide both sides by $sqrt[3]3$. Then the left is $sqrt[3]1+a +sqrt[3]1-a$ (for an $a$ which is in $[0,1)$) and we want to show that this is strictly less than $2$.
Let $f(x) = sqrt[3]1+x +sqrt[3]1-x$. Then $f(0)=2$. And it can be easily shown that $f'(x) < 0$ for all $xin (0,1)$ (since $(1+x)^frac23>(1-x)^frac23$ for all $x$ on that interval). Hence $f$ is strictly decreasing on $(0, 1)$, and therefore strictly less than 2.
answered 17 mins ago
bloomers
7261210
edited 10 mins ago
answered 17 mins ago
bloomers
7261210
answered 17 mins ago
bloomers
7261210
answered 17 mins ago
bloomers
7261210
7261210
add a comment |Â
add a comment |Â
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