Mixing
Find min of $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b}+4sqrt{2}sqrt{frac{ab+bc+ac}{a^2+b^2+c^2}}$ [closed]
$begingroup$
Given the three real numbers a, b, c are not negative, in which at most some are equal to zero.
Find min of $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b}+4sqrt{2}sqrt{frac{ab+bc+ac}{a^2+b^2+c^2}}$
Thanks so much
radicals substitution a.m.-g.m.-inequality uvw
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
$endgroup$
closed as off-topic by RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn Jan 28 at 16:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given the three real numbers a, b, c are not negative, in which at most some are equal to zero.
Find min of $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b}+4sqrt{2}sqrt{frac{ab+bc+ac}{a^2+b^2+c^2}}$
Thanks so much
radicals substitution a.m.-g.m.-inequality uvw
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
$endgroup$
closed as off-topic by RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn Jan 28 at 16:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Using $a=b=c$ gets me to $frac 12+frac 12+frac 12+4sqrt 2 cdot sqrt 1=4sqrt 2+frac 32.$ Can you check to see if there is a value smaller than this?
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:22
$begingroup$
Oh yes but I can't prove
$endgroup$
– Trương Văn Hào
Jan 28 at 6:22
$begingroup$
Setting $a=b$ and $c=0$ gives me $1+1+0+4sqrt 2 cdot sqrt {frac 12}=2+4=6 lt 4sqrt 2+frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:26
$begingroup$
Can you prove it $geq$ 6 ?? I think you correct .
$endgroup$
– Trương Văn Hào
Jan 28 at 6:28
$begingroup$
I think you meant that the $text {min} leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:37
add a comment |
$begingroup$
Given the three real numbers a, b, c are not negative, in which at most some are equal to zero.
Find min of $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b}+4sqrt{2}sqrt{frac{ab+bc+ac}{a^2+b^2+c^2}}$
Thanks so much
radicals substitution a.m.-g.m.-inequality uvw
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
$endgroup$
Given the three real numbers a, b, c are not negative, in which at most some are equal to zero.
Find min of $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b}+4sqrt{2}sqrt{frac{ab+bc+ac}{a^2+b^2+c^2}}$
Thanks so much
radicals substitution a.m.-g.m.-inequality uvw
radicals substitution a.m.-g.m.-inequality uvw
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
edited Jan 28 at 7:03
Michael Rozenberg
109k1896201
109k1896201
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
asked Jan 28 at 6:12
Trương Văn HàoTrương Văn Hào
51
51
closed as off-topic by RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn Jan 28 at 16:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn Jan 28 at 16:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Using $a=b=c$ gets me to $frac 12+frac 12+frac 12+4sqrt 2 cdot sqrt 1=4sqrt 2+frac 32.$ Can you check to see if there is a value smaller than this?
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:22
$begingroup$
Oh yes but I can't prove
$endgroup$
– Trương Văn Hào
Jan 28 at 6:22
$begingroup$
Setting $a=b$ and $c=0$ gives me $1+1+0+4sqrt 2 cdot sqrt {frac 12}=2+4=6 lt 4sqrt 2+frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:26
$begingroup$
Can you prove it $geq$ 6 ?? I think you correct .
$endgroup$
– Trương Văn Hào
Jan 28 at 6:28
$begingroup$
I think you meant that the $text {min} leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:37
add a comment |
$begingroup$
Using $a=b=c$ gets me to $frac 12+frac 12+frac 12+4sqrt 2 cdot sqrt 1=4sqrt 2+frac 32.$ Can you check to see if there is a value smaller than this?
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:22
$begingroup$
Oh yes but I can't prove
$endgroup$
– Trương Văn Hào
Jan 28 at 6:22
$begingroup$
Setting $a=b$ and $c=0$ gives me $1+1+0+4sqrt 2 cdot sqrt {frac 12}=2+4=6 lt 4sqrt 2+frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:26
$begingroup$
Can you prove it $geq$ 6 ?? I think you correct .
$endgroup$
– Trương Văn Hào
Jan 28 at 6:28
$begingroup$
I think you meant that the $text {min} leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:37
$begingroup$
Using $a=b=c$ gets me to $frac 12+frac 12+frac 12+4sqrt 2 cdot sqrt 1=4sqrt 2+frac 32.$ Can you check to see if there is a value smaller than this?
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:22
$begingroup$
Using $a=b=c$ gets me to $frac 12+frac 12+frac 12+4sqrt 2 cdot sqrt 1=4sqrt 2+frac 32.$ Can you check to see if there is a value smaller than this?
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:22
$begingroup$
Oh yes but I can't prove
$endgroup$
– Trương Văn Hào
Jan 28 at 6:22
$begingroup$
Oh yes but I can't prove
$endgroup$
– Trương Văn Hào
Jan 28 at 6:22
$begingroup$
Setting $a=b$ and $c=0$ gives me $1+1+0+4sqrt 2 cdot sqrt {frac 12}=2+4=6 lt 4sqrt 2+frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:26
$begingroup$
Setting $a=b$ and $c=0$ gives me $1+1+0+4sqrt 2 cdot sqrt {frac 12}=2+4=6 lt 4sqrt 2+frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:26
$begingroup$
Can you prove it $geq$ 6 ?? I think you correct .
$endgroup$
– Trương Văn Hào
Jan 28 at 6:28
$begingroup$
Can you prove it $geq$ 6 ?? I think you correct .
$endgroup$
– Trương Văn Hào
Jan 28 at 6:28
$begingroup$
I think you meant that the $text {min} leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:37
$begingroup$
I think you meant that the $text {min} leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $a=b=1$ and $c=0$ we get a value $6$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$sum_{cyc}frac{a}{b+c}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{sumlimits_{cyc}(a^3+a^2b+a^2c+abc)}{prodlimits_{cyc}(a+b)}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{3u(9u^2-6v^2)+3w^3}{9uv^2-w^3}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6,$$ which is $f(w^3)geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$,
which happens in the following cases.
- Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that
$$frac{a}{2}+frac{2}{a+1}+4sqrt{frac{2(2a+1)}{a^2+2}}geq6,$$ which is smooth;
$b=1$ and $c=0$.
We need to prove that
$$a+frac{1}{a}+4sqrt2sqrt{frac{a}{a^2+1}}geq6,$$ which is smooth too:
$$a+frac{1}{a}-2geq4left(1-sqrt{frac{2a}{a^2+1}}right)$$ or
$$frac{1}{a}geqfrac{4}{sqrt{a^2+1}left(sqrt{a^2+1}+sqrt{2a}right)}$$ or
$$a^2+1+sqrt{2a(a^2+1)}geq4a,$$ which is true by AM-GM:
$$a^2+1+sqrt{2a(a^2+1)}geq2a+sqrt{2acdot2a}=4a.$$
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
add a comment |
$begingroup$
Setting $c=0$, we find that it has simplified to finding $text{min } frac ab+ frac ba +4sqrt 2 sqrt {frac{ab}{a^2+b^2}}$
Assuming $a geq b$, we set $a=kb$ to obtain $k+frac 1k+4sqrt 2 sqrt {frac {k}{k^2+1}}=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$
As we have to find $text {min} frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}},$ we could try and figure out the minimum of $f(k)=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$ by finding $k$ when $f'(k)=0.$ However, being not skilled enough in that aspect, I opted to graph it and find the minimum point.
A we can see, the minimum of $f(k)=6$ occurs when $k=1,$ giving us a minimum when $a=b$ and $c=0.$
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
1
$begingroup$
Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
$endgroup$
– TheSimpliFire
Mar 15 at 7:17
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $a=b=1$ and $c=0$ we get a value $6$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$sum_{cyc}frac{a}{b+c}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{sumlimits_{cyc}(a^3+a^2b+a^2c+abc)}{prodlimits_{cyc}(a+b)}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{3u(9u^2-6v^2)+3w^3}{9uv^2-w^3}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6,$$ which is $f(w^3)geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$,
which happens in the following cases.
- Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that
$$frac{a}{2}+frac{2}{a+1}+4sqrt{frac{2(2a+1)}{a^2+2}}geq6,$$ which is smooth;
$b=1$ and $c=0$.
We need to prove that
$$a+frac{1}{a}+4sqrt2sqrt{frac{a}{a^2+1}}geq6,$$ which is smooth too:
$$a+frac{1}{a}-2geq4left(1-sqrt{frac{2a}{a^2+1}}right)$$ or
$$frac{1}{a}geqfrac{4}{sqrt{a^2+1}left(sqrt{a^2+1}+sqrt{2a}right)}$$ or
$$a^2+1+sqrt{2a(a^2+1)}geq4a,$$ which is true by AM-GM:
$$a^2+1+sqrt{2a(a^2+1)}geq2a+sqrt{2acdot2a}=4a.$$
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
add a comment |
$begingroup$
For $a=b=1$ and $c=0$ we get a value $6$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$sum_{cyc}frac{a}{b+c}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{sumlimits_{cyc}(a^3+a^2b+a^2c+abc)}{prodlimits_{cyc}(a+b)}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{3u(9u^2-6v^2)+3w^3}{9uv^2-w^3}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6,$$ which is $f(w^3)geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$,
which happens in the following cases.
- Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that
$$frac{a}{2}+frac{2}{a+1}+4sqrt{frac{2(2a+1)}{a^2+2}}geq6,$$ which is smooth;
$b=1$ and $c=0$.
We need to prove that
$$a+frac{1}{a}+4sqrt2sqrt{frac{a}{a^2+1}}geq6,$$ which is smooth too:
$$a+frac{1}{a}-2geq4left(1-sqrt{frac{2a}{a^2+1}}right)$$ or
$$frac{1}{a}geqfrac{4}{sqrt{a^2+1}left(sqrt{a^2+1}+sqrt{2a}right)}$$ or
$$a^2+1+sqrt{2a(a^2+1)}geq4a,$$ which is true by AM-GM:
$$a^2+1+sqrt{2a(a^2+1)}geq2a+sqrt{2acdot2a}=4a.$$
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
add a comment |
$begingroup$
For $a=b=1$ and $c=0$ we get a value $6$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$sum_{cyc}frac{a}{b+c}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{sumlimits_{cyc}(a^3+a^2b+a^2c+abc)}{prodlimits_{cyc}(a+b)}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{3u(9u^2-6v^2)+3w^3}{9uv^2-w^3}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6,$$ which is $f(w^3)geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$,
which happens in the following cases.
- Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that
$$frac{a}{2}+frac{2}{a+1}+4sqrt{frac{2(2a+1)}{a^2+2}}geq6,$$ which is smooth;
$b=1$ and $c=0$.
We need to prove that
$$a+frac{1}{a}+4sqrt2sqrt{frac{a}{a^2+1}}geq6,$$ which is smooth too:
$$a+frac{1}{a}-2geq4left(1-sqrt{frac{2a}{a^2+1}}right)$$ or
$$frac{1}{a}geqfrac{4}{sqrt{a^2+1}left(sqrt{a^2+1}+sqrt{2a}right)}$$ or
$$a^2+1+sqrt{2a(a^2+1)}geq4a,$$ which is true by AM-GM:
$$a^2+1+sqrt{2a(a^2+1)}geq2a+sqrt{2acdot2a}=4a.$$
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
For $a=b=1$ and $c=0$ we get a value $6$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$sum_{cyc}frac{a}{b+c}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{sumlimits_{cyc}(a^3+a^2b+a^2c+abc)}{prodlimits_{cyc}(a+b)}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6$$ or
$$frac{3u(9u^2-6v^2)+3w^3}{9uv^2-w^3}+4sqrt{frac{2v^2}{3u^2-2v^2}}geq6,$$ which is $f(w^3)geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$,
which happens in the following cases.
- Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that
$$frac{a}{2}+frac{2}{a+1}+4sqrt{frac{2(2a+1)}{a^2+2}}geq6,$$ which is smooth;
$b=1$ and $c=0$.
We need to prove that
$$a+frac{1}{a}+4sqrt2sqrt{frac{a}{a^2+1}}geq6,$$ which is smooth too:
$$a+frac{1}{a}-2geq4left(1-sqrt{frac{2a}{a^2+1}}right)$$ or
$$frac{1}{a}geqfrac{4}{sqrt{a^2+1}left(sqrt{a^2+1}+sqrt{2a}right)}$$ or
$$a^2+1+sqrt{2a(a^2+1)}geq4a,$$ which is true by AM-GM:
$$a^2+1+sqrt{2a(a^2+1)}geq2a+sqrt{2acdot2a}=4a.$$
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
edited Jan 28 at 6:56
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 6:51
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
add a comment |
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
Shouldn't it be true for all $a=b gt 0$ and $c=0?$
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 7:02
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
$begingroup$
@Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case.
$endgroup$
– Michael Rozenberg
Jan 28 at 7:04
add a comment |
$begingroup$
Setting $c=0$, we find that it has simplified to finding $text{min } frac ab+ frac ba +4sqrt 2 sqrt {frac{ab}{a^2+b^2}}$
Assuming $a geq b$, we set $a=kb$ to obtain $k+frac 1k+4sqrt 2 sqrt {frac {k}{k^2+1}}=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$
As we have to find $text {min} frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}},$ we could try and figure out the minimum of $f(k)=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$ by finding $k$ when $f'(k)=0.$ However, being not skilled enough in that aspect, I opted to graph it and find the minimum point.
A we can see, the minimum of $f(k)=6$ occurs when $k=1,$ giving us a minimum when $a=b$ and $c=0.$
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
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1
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Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
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– TheSimpliFire
Mar 15 at 7:17
add a comment |
$begingroup$
Setting $c=0$, we find that it has simplified to finding $text{min } frac ab+ frac ba +4sqrt 2 sqrt {frac{ab}{a^2+b^2}}$
Assuming $a geq b$, we set $a=kb$ to obtain $k+frac 1k+4sqrt 2 sqrt {frac {k}{k^2+1}}=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$
As we have to find $text {min} frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}},$ we could try and figure out the minimum of $f(k)=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$ by finding $k$ when $f'(k)=0.$ However, being not skilled enough in that aspect, I opted to graph it and find the minimum point.
A we can see, the minimum of $f(k)=6$ occurs when $k=1,$ giving us a minimum when $a=b$ and $c=0.$
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
1
$begingroup$
Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
$endgroup$
– TheSimpliFire
Mar 15 at 7:17
add a comment |
$begingroup$
Setting $c=0$, we find that it has simplified to finding $text{min } frac ab+ frac ba +4sqrt 2 sqrt {frac{ab}{a^2+b^2}}$
Assuming $a geq b$, we set $a=kb$ to obtain $k+frac 1k+4sqrt 2 sqrt {frac {k}{k^2+1}}=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$
As we have to find $text {min} frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}},$ we could try and figure out the minimum of $f(k)=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$ by finding $k$ when $f'(k)=0.$ However, being not skilled enough in that aspect, I opted to graph it and find the minimum point.
A we can see, the minimum of $f(k)=6$ occurs when $k=1,$ giving us a minimum when $a=b$ and $c=0.$
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
Setting $c=0$, we find that it has simplified to finding $text{min } frac ab+ frac ba +4sqrt 2 sqrt {frac{ab}{a^2+b^2}}$
Assuming $a geq b$, we set $a=kb$ to obtain $k+frac 1k+4sqrt 2 sqrt {frac {k}{k^2+1}}=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$
As we have to find $text {min} frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}},$ we could try and figure out the minimum of $f(k)=frac {k^2+1}k+4sqrt 2 sqrt {frac k{k^2+1}}$ by finding $k$ when $f'(k)=0.$ However, being not skilled enough in that aspect, I opted to graph it and find the minimum point.
A we can see, the minimum of $f(k)=6$ occurs when $k=1,$ giving us a minimum when $a=b$ and $c=0.$
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
edited Jan 28 at 7:01
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
answered Jan 28 at 6:50
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
1,6792625
1
$begingroup$
Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
$endgroup$
– TheSimpliFire
Mar 15 at 7:17
add a comment |
1
$begingroup$
Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
$endgroup$
– TheSimpliFire
Mar 15 at 7:17
1
1
$begingroup$
Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
$endgroup$
– TheSimpliFire
Mar 15 at 7:17
$begingroup$
Note that $$f'(k)=1-frac1{k^2}-frac{2sqrt2}{sqrt{frac{k}{k^2+1}}}cdotfrac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0impliesfrac{2sqrt2(k^2-1)}{(k^2+1)^2}sqrt{frac{k^2+1}k}=1-frac1{k^2}$$ giving $$frac{k^2-1}{(k^2+1)^2}sqrt{frac{k^2+1}k}=frac1{2sqrt2}left(1-frac1{k^2}right)implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$.
$endgroup$
– TheSimpliFire
Mar 15 at 7:17
add a comment |
$begingroup$
Using $a=b=c$ gets me to $frac 12+frac 12+frac 12+4sqrt 2 cdot sqrt 1=4sqrt 2+frac 32.$ Can you check to see if there is a value smaller than this?
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– Mohammad Zuhair Khan
Jan 28 at 6:22
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Oh yes but I can't prove
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– Trương Văn Hào
Jan 28 at 6:22
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Setting $a=b$ and $c=0$ gives me $1+1+0+4sqrt 2 cdot sqrt {frac 12}=2+4=6 lt 4sqrt 2+frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$
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– Mohammad Zuhair Khan
Jan 28 at 6:26
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Can you prove it $geq$ 6 ?? I think you correct .
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– Trương Văn Hào
Jan 28 at 6:28
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I think you meant that the $text {min} leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof.
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– Mohammad Zuhair Khan
Jan 28 at 6:37