Mixing
Generalization of Nesbitts's inequality
$begingroup$
Let some (fixed) real $k >0$ and positive reals $a,b,c$. Consider the conjecture
$$
left(frac{a}{b+c}right)^k +left(frac{b}{a+c}right)^k+left(frac{c}{a+b}right)^k geq min {frac{3}{2^k} ; 2 }
$$
Some cases are known:
$k = frac12$ gives the bound 2, which is proved here.
$k = 1$ gives the bound 1.5, which is Nesbitt's inequality
I guess $k = 2$ should be known but I couldn't find it.
Does the conjecture hold?
algebra-precalculus inequality jensen-inequality karamata-inequality
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
$endgroup$
add a comment |
$begingroup$
Let some (fixed) real $k >0$ and positive reals $a,b,c$. Consider the conjecture
$$
left(frac{a}{b+c}right)^k +left(frac{b}{a+c}right)^k+left(frac{c}{a+b}right)^k geq min {frac{3}{2^k} ; 2 }
$$
Some cases are known:
$k = frac12$ gives the bound 2, which is proved here.
$k = 1$ gives the bound 1.5, which is Nesbitt's inequality
I guess $k = 2$ should be known but I couldn't find it.
Does the conjecture hold?
algebra-precalculus inequality jensen-inequality karamata-inequality
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
$endgroup$
$begingroup$
see here uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_18.pdf
$endgroup$
– Dr. Sonnhard Graubner
Feb 1 at 18:19
$begingroup$
There is a very nice proof of your inequality for $k=frac{27}{46}.$
$endgroup$
– Michael Rozenberg
Feb 1 at 18:31
$begingroup$
@MichaelRozenberg "There is ... " ?
$endgroup$
– Andreas
Feb 1 at 19:31
$begingroup$
Yes, I found this proof.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:44
add a comment |
$begingroup$
Let some (fixed) real $k >0$ and positive reals $a,b,c$. Consider the conjecture
$$
left(frac{a}{b+c}right)^k +left(frac{b}{a+c}right)^k+left(frac{c}{a+b}right)^k geq min {frac{3}{2^k} ; 2 }
$$
Some cases are known:
$k = frac12$ gives the bound 2, which is proved here.
$k = 1$ gives the bound 1.5, which is Nesbitt's inequality
I guess $k = 2$ should be known but I couldn't find it.
Does the conjecture hold?
algebra-precalculus inequality jensen-inequality karamata-inequality
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
$endgroup$
Let some (fixed) real $k >0$ and positive reals $a,b,c$. Consider the conjecture
$$
left(frac{a}{b+c}right)^k +left(frac{b}{a+c}right)^k+left(frac{c}{a+b}right)^k geq min {frac{3}{2^k} ; 2 }
$$
Some cases are known:
$k = frac12$ gives the bound 2, which is proved here.
$k = 1$ gives the bound 1.5, which is Nesbitt's inequality
I guess $k = 2$ should be known but I couldn't find it.
Does the conjecture hold?
algebra-precalculus inequality jensen-inequality karamata-inequality
algebra-precalculus inequality jensen-inequality karamata-inequality
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
edited Feb 4 at 8:54
Andreas
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
asked Feb 1 at 18:09
AndreasAndreas
8,4411137
8,4411137
$begingroup$
see here uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_18.pdf
$endgroup$
– Dr. Sonnhard Graubner
Feb 1 at 18:19
$begingroup$
There is a very nice proof of your inequality for $k=frac{27}{46}.$
$endgroup$
– Michael Rozenberg
Feb 1 at 18:31
$begingroup$
@MichaelRozenberg "There is ... " ?
$endgroup$
– Andreas
Feb 1 at 19:31
$begingroup$
Yes, I found this proof.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:44
add a comment |
$begingroup$
see here uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_18.pdf
$endgroup$
– Dr. Sonnhard Graubner
Feb 1 at 18:19
$begingroup$
There is a very nice proof of your inequality for $k=frac{27}{46}.$
$endgroup$
– Michael Rozenberg
Feb 1 at 18:31
$begingroup$
@MichaelRozenberg "There is ... " ?
$endgroup$
– Andreas
Feb 1 at 19:31
$begingroup$
Yes, I found this proof.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:44
$begingroup$
see here uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_18.pdf
$endgroup$
– Dr. Sonnhard Graubner
Feb 1 at 18:19
$begingroup$
see here uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_18.pdf
$endgroup$
– Dr. Sonnhard Graubner
Feb 1 at 18:19
$begingroup$
There is a very nice proof of your inequality for $k=frac{27}{46}.$
$endgroup$
– Michael Rozenberg
Feb 1 at 18:31
$begingroup$
There is a very nice proof of your inequality for $k=frac{27}{46}.$
$endgroup$
– Michael Rozenberg
Feb 1 at 18:31
$begingroup$
@MichaelRozenberg "There is ... " ?
$endgroup$
– Andreas
Feb 1 at 19:31
$begingroup$
@MichaelRozenberg "There is ... " ?
$endgroup$
– Andreas
Feb 1 at 19:31
$begingroup$
Yes, I found this proof.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:44
$begingroup$
Yes, I found this proof.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1.$frac{3}{2^k}leq2$ or $kgeqlog_23-1=0.5849...$.
Now, let $a+b+c=3$.
Thus, we need to prove that $sumlimits_{cyc}f(a)geqfrac{3}{2^k}$, where $$f(x)=left(frac{x}{3-x}right)^k.$$
Indeed, $$f''(x)=frac{3kx^{k-2}(2x+3k-3)}{(3-x)^{k+2}}>0$$ for all $1leq x<3,$ which says that by Vasc's RCF Theorem it's enough to check our inequality for an equality case of two variables, which is smooth enough.
About RCF Theorem see here:
https://pdfs.semanticscholar.org/1419/e0baa6b073c5903e430ddb0c9a154c61d208.pdf
- $0<kleqfrac{1}{2}.$
We need to prove that
$$sum_{cyc}left(frac{a}{b+c}right)^kgeq2.$$
Indeed, let $a^k=sqrt{x},$ $b^k=sqrt{y}$ and $c^k=sqrt{z}.$
Thus, since $g(x)=x^{frac{1}{2k}}$ is a convex function, by Karamata we obtain:
$$(y+z)^{frac{1}{2k}}+0geq y^{frac{1}{2k}}+z^{frac{1}{2k}},$$ which gives $$sum_{cyc}left(frac{a}{b+c}right)^k=sum_{cyc}frac{sqrt{x}}{left(y^{frac{1}{2k}}+z^{frac{1}{2k}}right)^k}geqsum_{cyc}sqrt{frac{x}{y+z}}geq2.$$
- $frac{1}{2}<k<log_2frac{3}{2}.$
In this case I have no a nice proof.
Since $f$ has on $(0,3)$ an unique inflection point for $x=x_0=frac{3}{2}(1-k)<1,$
it's impossible that ${a,b,c}subsetleft(0,x_0right).$
Now, let $aleq bleq c$.
We have three cases.
a) $aleq bleq x_0leq c.$
In this case we can use Karamata and we'll get an inequality of one variable.
b) $aleq x_0leq bleq c$.
In this case we can use Jensen and we'll get an inequality of one variable.
c) $x_0leq aleq bleq c$.
In this case our inequality is obviously true by Jensen.
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
add a comment |
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$begingroup$
1.$frac{3}{2^k}leq2$ or $kgeqlog_23-1=0.5849...$.
Now, let $a+b+c=3$.
Thus, we need to prove that $sumlimits_{cyc}f(a)geqfrac{3}{2^k}$, where $$f(x)=left(frac{x}{3-x}right)^k.$$
Indeed, $$f''(x)=frac{3kx^{k-2}(2x+3k-3)}{(3-x)^{k+2}}>0$$ for all $1leq x<3,$ which says that by Vasc's RCF Theorem it's enough to check our inequality for an equality case of two variables, which is smooth enough.
About RCF Theorem see here:
https://pdfs.semanticscholar.org/1419/e0baa6b073c5903e430ddb0c9a154c61d208.pdf
- $0<kleqfrac{1}{2}.$
We need to prove that
$$sum_{cyc}left(frac{a}{b+c}right)^kgeq2.$$
Indeed, let $a^k=sqrt{x},$ $b^k=sqrt{y}$ and $c^k=sqrt{z}.$
Thus, since $g(x)=x^{frac{1}{2k}}$ is a convex function, by Karamata we obtain:
$$(y+z)^{frac{1}{2k}}+0geq y^{frac{1}{2k}}+z^{frac{1}{2k}},$$ which gives $$sum_{cyc}left(frac{a}{b+c}right)^k=sum_{cyc}frac{sqrt{x}}{left(y^{frac{1}{2k}}+z^{frac{1}{2k}}right)^k}geqsum_{cyc}sqrt{frac{x}{y+z}}geq2.$$
- $frac{1}{2}<k<log_2frac{3}{2}.$
In this case I have no a nice proof.
Since $f$ has on $(0,3)$ an unique inflection point for $x=x_0=frac{3}{2}(1-k)<1,$
it's impossible that ${a,b,c}subsetleft(0,x_0right).$
Now, let $aleq bleq c$.
We have three cases.
a) $aleq bleq x_0leq c.$
In this case we can use Karamata and we'll get an inequality of one variable.
b) $aleq x_0leq bleq c$.
In this case we can use Jensen and we'll get an inequality of one variable.
c) $x_0leq aleq bleq c$.
In this case our inequality is obviously true by Jensen.
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
add a comment |
$begingroup$
1.$frac{3}{2^k}leq2$ or $kgeqlog_23-1=0.5849...$.
Now, let $a+b+c=3$.
Thus, we need to prove that $sumlimits_{cyc}f(a)geqfrac{3}{2^k}$, where $$f(x)=left(frac{x}{3-x}right)^k.$$
Indeed, $$f''(x)=frac{3kx^{k-2}(2x+3k-3)}{(3-x)^{k+2}}>0$$ for all $1leq x<3,$ which says that by Vasc's RCF Theorem it's enough to check our inequality for an equality case of two variables, which is smooth enough.
About RCF Theorem see here:
https://pdfs.semanticscholar.org/1419/e0baa6b073c5903e430ddb0c9a154c61d208.pdf
- $0<kleqfrac{1}{2}.$
We need to prove that
$$sum_{cyc}left(frac{a}{b+c}right)^kgeq2.$$
Indeed, let $a^k=sqrt{x},$ $b^k=sqrt{y}$ and $c^k=sqrt{z}.$
Thus, since $g(x)=x^{frac{1}{2k}}$ is a convex function, by Karamata we obtain:
$$(y+z)^{frac{1}{2k}}+0geq y^{frac{1}{2k}}+z^{frac{1}{2k}},$$ which gives $$sum_{cyc}left(frac{a}{b+c}right)^k=sum_{cyc}frac{sqrt{x}}{left(y^{frac{1}{2k}}+z^{frac{1}{2k}}right)^k}geqsum_{cyc}sqrt{frac{x}{y+z}}geq2.$$
- $frac{1}{2}<k<log_2frac{3}{2}.$
In this case I have no a nice proof.
Since $f$ has on $(0,3)$ an unique inflection point for $x=x_0=frac{3}{2}(1-k)<1,$
it's impossible that ${a,b,c}subsetleft(0,x_0right).$
Now, let $aleq bleq c$.
We have three cases.
a) $aleq bleq x_0leq c.$
In this case we can use Karamata and we'll get an inequality of one variable.
b) $aleq x_0leq bleq c$.
In this case we can use Jensen and we'll get an inequality of one variable.
c) $x_0leq aleq bleq c$.
In this case our inequality is obviously true by Jensen.
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
add a comment |
$begingroup$
1.$frac{3}{2^k}leq2$ or $kgeqlog_23-1=0.5849...$.
Now, let $a+b+c=3$.
Thus, we need to prove that $sumlimits_{cyc}f(a)geqfrac{3}{2^k}$, where $$f(x)=left(frac{x}{3-x}right)^k.$$
Indeed, $$f''(x)=frac{3kx^{k-2}(2x+3k-3)}{(3-x)^{k+2}}>0$$ for all $1leq x<3,$ which says that by Vasc's RCF Theorem it's enough to check our inequality for an equality case of two variables, which is smooth enough.
About RCF Theorem see here:
https://pdfs.semanticscholar.org/1419/e0baa6b073c5903e430ddb0c9a154c61d208.pdf
- $0<kleqfrac{1}{2}.$
We need to prove that
$$sum_{cyc}left(frac{a}{b+c}right)^kgeq2.$$
Indeed, let $a^k=sqrt{x},$ $b^k=sqrt{y}$ and $c^k=sqrt{z}.$
Thus, since $g(x)=x^{frac{1}{2k}}$ is a convex function, by Karamata we obtain:
$$(y+z)^{frac{1}{2k}}+0geq y^{frac{1}{2k}}+z^{frac{1}{2k}},$$ which gives $$sum_{cyc}left(frac{a}{b+c}right)^k=sum_{cyc}frac{sqrt{x}}{left(y^{frac{1}{2k}}+z^{frac{1}{2k}}right)^k}geqsum_{cyc}sqrt{frac{x}{y+z}}geq2.$$
- $frac{1}{2}<k<log_2frac{3}{2}.$
In this case I have no a nice proof.
Since $f$ has on $(0,3)$ an unique inflection point for $x=x_0=frac{3}{2}(1-k)<1,$
it's impossible that ${a,b,c}subsetleft(0,x_0right).$
Now, let $aleq bleq c$.
We have three cases.
a) $aleq bleq x_0leq c.$
In this case we can use Karamata and we'll get an inequality of one variable.
b) $aleq x_0leq bleq c$.
In this case we can use Jensen and we'll get an inequality of one variable.
c) $x_0leq aleq bleq c$.
In this case our inequality is obviously true by Jensen.
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
1.$frac{3}{2^k}leq2$ or $kgeqlog_23-1=0.5849...$.
Now, let $a+b+c=3$.
Thus, we need to prove that $sumlimits_{cyc}f(a)geqfrac{3}{2^k}$, where $$f(x)=left(frac{x}{3-x}right)^k.$$
Indeed, $$f''(x)=frac{3kx^{k-2}(2x+3k-3)}{(3-x)^{k+2}}>0$$ for all $1leq x<3,$ which says that by Vasc's RCF Theorem it's enough to check our inequality for an equality case of two variables, which is smooth enough.
About RCF Theorem see here:
https://pdfs.semanticscholar.org/1419/e0baa6b073c5903e430ddb0c9a154c61d208.pdf
- $0<kleqfrac{1}{2}.$
We need to prove that
$$sum_{cyc}left(frac{a}{b+c}right)^kgeq2.$$
Indeed, let $a^k=sqrt{x},$ $b^k=sqrt{y}$ and $c^k=sqrt{z}.$
Thus, since $g(x)=x^{frac{1}{2k}}$ is a convex function, by Karamata we obtain:
$$(y+z)^{frac{1}{2k}}+0geq y^{frac{1}{2k}}+z^{frac{1}{2k}},$$ which gives $$sum_{cyc}left(frac{a}{b+c}right)^k=sum_{cyc}frac{sqrt{x}}{left(y^{frac{1}{2k}}+z^{frac{1}{2k}}right)^k}geqsum_{cyc}sqrt{frac{x}{y+z}}geq2.$$
- $frac{1}{2}<k<log_2frac{3}{2}.$
In this case I have no a nice proof.
Since $f$ has on $(0,3)$ an unique inflection point for $x=x_0=frac{3}{2}(1-k)<1,$
it's impossible that ${a,b,c}subsetleft(0,x_0right).$
Now, let $aleq bleq c$.
We have three cases.
a) $aleq bleq x_0leq c.$
In this case we can use Karamata and we'll get an inequality of one variable.
b) $aleq x_0leq bleq c$.
In this case we can use Jensen and we'll get an inequality of one variable.
c) $x_0leq aleq bleq c$.
In this case our inequality is obviously true by Jensen.
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
edited Feb 1 at 22:53
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 18:26
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
add a comment |
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
You consider the square root ($k = frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ?
$endgroup$
– Andreas
Feb 1 at 19:30
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
$begingroup$
@Andreas Sorry. I fixed.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:55
add a comment |
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$begingroup$
see here uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_18.pdf
$endgroup$
– Dr. Sonnhard Graubner
Feb 1 at 18:19
$begingroup$
There is a very nice proof of your inequality for $k=frac{27}{46}.$
$endgroup$
– Michael Rozenberg
Feb 1 at 18:31
$begingroup$
@MichaelRozenberg "There is ... " ?
$endgroup$
– Andreas
Feb 1 at 19:31
$begingroup$
Yes, I found this proof.
$endgroup$
– Michael Rozenberg
Feb 1 at 20:44