Mixing
![Given d/dx (f(x)) = sqrt f(x), find d/dx ( sqrt f(x)). [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Inequality with $abc=2$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $fraca+b+c4geq sqrt[4]fraca^2+b^2+c^26$
Sorry, I don't have an idea for this problem.If I had, I would have showed them all.
We need to prove that:
$$left(fraca+b+c4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24.$$
Now, by the Vasc's inequality
$(a+b+c)^5geq81abc(a^2+b^2+c^2)$
We can obtain
$$a+b+cgeqsqrt[5]81abc(a^2+b^2+c^2)$$
so it's enough to prove that
$$left(fracsqrt[5]81abc(a^2+b^2+c^2)4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24,$$
which is wrong for $abcrightarrow0^+$.
real-analysis inequality logarithms substitution uvw
edited yesterday
Michael Rozenberg
86.9k1575178
asked yesterday
Tsukuyomi
62
add a comment |Â
up vote
1
down vote
favorite
Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $fraca+b+c4geq sqrt[4]fraca^2+b^2+c^26$
Sorry, I don't have an idea for this problem.If I had, I would have showed them all.
We need to prove that:
$$left(fraca+b+c4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24.$$
Now, by the Vasc's inequality
$(a+b+c)^5geq81abc(a^2+b^2+c^2)$
We can obtain
$$a+b+cgeqsqrt[5]81abc(a^2+b^2+c^2)$$
so it's enough to prove that
$$left(fracsqrt[5]81abc(a^2+b^2+c^2)4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24,$$
which is wrong for $abcrightarrow0^+$.
real-analysis inequality logarithms substitution uvw
edited yesterday
Michael Rozenberg
86.9k1575178
asked yesterday
Tsukuyomi
62
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
yesterday
You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. )
– Cataline
21 hours ago
@Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$
– Michael Rozenberg
5 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $fraca+b+c4geq sqrt[4]fraca^2+b^2+c^26$
Sorry, I don't have an idea for this problem.If I had, I would have showed them all.
We need to prove that:
$$left(fraca+b+c4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24.$$
Now, by the Vasc's inequality
$(a+b+c)^5geq81abc(a^2+b^2+c^2)$
We can obtain
$$a+b+cgeqsqrt[5]81abc(a^2+b^2+c^2)$$
so it's enough to prove that
$$left(fracsqrt[5]81abc(a^2+b^2+c^2)4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24,$$
which is wrong for $abcrightarrow0^+$.
real-analysis inequality logarithms substitution uvw
edited yesterday
Michael Rozenberg
86.9k1575178
asked yesterday
Tsukuyomi
62
Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $fraca+b+c4geq sqrt[4]fraca^2+b^2+c^26$
Sorry, I don't have an idea for this problem.If I had, I would have showed them all.
We need to prove that:
$$left(fraca+b+c4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24.$$
Now, by the Vasc's inequality
$(a+b+c)^5geq81abc(a^2+b^2+c^2)$
We can obtain
$$a+b+cgeqsqrt[5]81abc(a^2+b^2+c^2)$$
so it's enough to prove that
$$left(fracsqrt[5]81abc(a^2+b^2+c^2)4right)^12geqleft(fraca^2+b^2+c^26right)^3cdotfraca^2b^2c^24,$$
which is wrong for $abcrightarrow0^+$.
real-analysis inequality logarithms substitution uvw
edited yesterday
Michael Rozenberg
86.9k1575178
asked yesterday
Tsukuyomi
62
edited yesterday
Michael Rozenberg
86.9k1575178
edited yesterday
Michael Rozenberg
86.9k1575178
edited yesterday
Michael Rozenberg
86.9k1575178
86.9k1575178
asked yesterday
Tsukuyomi
62
asked yesterday
Tsukuyomi
62
asked yesterday
Tsukuyomi
62
62
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
yesterday
You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. )
– Cataline
21 hours ago
@Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$
– Michael Rozenberg
5 hours ago
add a comment |Â
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
yesterday
You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. )
– Cataline
21 hours ago
@Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$
– Michael Rozenberg
5 hours ago
1
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
yesterday
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
yesterday
You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. )
– Cataline
21 hours ago
You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. )
– Cataline
21 hours ago
@Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$
– Michael Rozenberg
5 hours ago
@Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$
– Michael Rozenberg
5 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality is equivalent to $f(v^2)geq0$, where
$$f(u)=frac34u-left(frac3u^2-2v^22right)^frac14left(fracw^32right)^frac16.$$
We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.
Now, let $a=b=sqrt[3]t$.
Thus, $c=frac2sqrt[3]t^2$ and we need to prove that
$$frac2sqrt[3]t+frac2sqrt[3]t^24geqsqrt[4]frac2sqrt[3]t^2+frac4sqrt[3]t^46$$ or
$$fract+12geqsqrt[3]tsqrt[4]fract^2+23$$ or $g(t)geq0,$ where
$$g(t)=lnfract+12-frac13lnt-frac14lnfract^2+23.$$
But $$g'(t)=frac1t+1-frac13t-fract2(t^2+2)=frac(t-2)^2(t-1)6t(t+1)(t^2+2),$$
which gives $t_min=1$ and since $g(1)=0,$ we are done!
answered yesterday
Michael Rozenberg
86.9k1575178
4
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
3
Try to find something better!
– Michael Rozenberg
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality is equivalent to $f(v^2)geq0$, where
$$f(u)=frac34u-left(frac3u^2-2v^22right)^frac14left(fracw^32right)^frac16.$$
We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.
Now, let $a=b=sqrt[3]t$.
Thus, $c=frac2sqrt[3]t^2$ and we need to prove that
$$frac2sqrt[3]t+frac2sqrt[3]t^24geqsqrt[4]frac2sqrt[3]t^2+frac4sqrt[3]t^46$$ or
$$fract+12geqsqrt[3]tsqrt[4]fract^2+23$$ or $g(t)geq0,$ where
$$g(t)=lnfract+12-frac13lnt-frac14lnfract^2+23.$$
But $$g'(t)=frac1t+1-frac13t-fract2(t^2+2)=frac(t-2)^2(t-1)6t(t+1)(t^2+2),$$
which gives $t_min=1$ and since $g(1)=0,$ we are done!
answered yesterday
Michael Rozenberg
86.9k1575178
4
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
3
Try to find something better!
– Michael Rozenberg
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
add a comment |Â
up vote
0
down vote
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality is equivalent to $f(v^2)geq0$, where
$$f(u)=frac34u-left(frac3u^2-2v^22right)^frac14left(fracw^32right)^frac16.$$
We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.
Now, let $a=b=sqrt[3]t$.
Thus, $c=frac2sqrt[3]t^2$ and we need to prove that
$$frac2sqrt[3]t+frac2sqrt[3]t^24geqsqrt[4]frac2sqrt[3]t^2+frac4sqrt[3]t^46$$ or
$$fract+12geqsqrt[3]tsqrt[4]fract^2+23$$ or $g(t)geq0,$ where
$$g(t)=lnfract+12-frac13lnt-frac14lnfract^2+23.$$
But $$g'(t)=frac1t+1-frac13t-fract2(t^2+2)=frac(t-2)^2(t-1)6t(t+1)(t^2+2),$$
which gives $t_min=1$ and since $g(1)=0,$ we are done!
answered yesterday
Michael Rozenberg
86.9k1575178
4
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
3
Try to find something better!
– Michael Rozenberg
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality is equivalent to $f(v^2)geq0$, where
$$f(u)=frac34u-left(frac3u^2-2v^22right)^frac14left(fracw^32right)^frac16.$$
We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.
Now, let $a=b=sqrt[3]t$.
Thus, $c=frac2sqrt[3]t^2$ and we need to prove that
$$frac2sqrt[3]t+frac2sqrt[3]t^24geqsqrt[4]frac2sqrt[3]t^2+frac4sqrt[3]t^46$$ or
$$fract+12geqsqrt[3]tsqrt[4]fract^2+23$$ or $g(t)geq0,$ where
$$g(t)=lnfract+12-frac13lnt-frac14lnfract^2+23.$$
But $$g'(t)=frac1t+1-frac13t-fract2(t^2+2)=frac(t-2)^2(t-1)6t(t+1)(t^2+2),$$
which gives $t_min=1$ and since $g(1)=0,$ we are done!
answered yesterday
Michael Rozenberg
86.9k1575178
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality is equivalent to $f(v^2)geq0$, where
$$f(u)=frac34u-left(frac3u^2-2v^22right)^frac14left(fracw^32right)^frac16.$$
We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.
Now, let $a=b=sqrt[3]t$.
Thus, $c=frac2sqrt[3]t^2$ and we need to prove that
$$frac2sqrt[3]t+frac2sqrt[3]t^24geqsqrt[4]frac2sqrt[3]t^2+frac4sqrt[3]t^46$$ or
$$fract+12geqsqrt[3]tsqrt[4]fract^2+23$$ or $g(t)geq0,$ where
$$g(t)=lnfract+12-frac13lnt-frac14lnfract^2+23.$$
But $$g'(t)=frac1t+1-frac13t-fract2(t^2+2)=frac(t-2)^2(t-1)6t(t+1)(t^2+2),$$
which gives $t_min=1$ and since $g(1)=0,$ we are done!
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
86.9k1575178
4
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
3
Try to find something better!
– Michael Rozenberg
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
add a comment |Â
4
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
3
Try to find something better!
– Michael Rozenberg
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
4
4
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
Thank you very much. I think this is not a nice solution.
– Tsukuyomi
yesterday
3
3
Try to find something better!
– Michael Rozenberg
yesterday
Try to find something better!
– Michael Rozenberg
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
Yes, I will post my solution for you at P.M in aops
– Tsukuyomi
yesterday
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
I am sure that it's impossible to find something better than my solution. Good luck!
– Michael Rozenberg
23 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872781%2finequality-with-abc-2%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
yesterday
You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. )
– Cataline
21 hours ago
@Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$
– Michael Rozenberg
5 hours ago