An inequality to finalize a proof
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Related to this Inequality $sumlimits_{cyc}frac{a^3}{13a^2+5b^2}geqfrac{a+b+c}{18}$ and my second answer I have to prove this :
Let $a,b,c$ be real positive numbers then we have : $$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq frac{ab+bc+ca}{18^2}$$
My try :
With Chebychev's inequality we get :
$$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq left(frac{ab+bc+ca}{3}right)left(sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)right) $$
Remains to prove :
$$sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)geq frac{3}{18^2}$$
Or with the right substitution :
$$sum_{cyc}left(frac{x}{13x+5y}right)left(frac{y}{13y+5z}right)geq frac{3}{18^2}$$
And after this I don't know what to do...
Can someone help me ?
Thanks
real-analysis inequality contest-math
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
asked Nov 23 at 8:01
max8128
1,031421
add a comment |
up vote
1
down vote
favorite
Related to this Inequality $sumlimits_{cyc}frac{a^3}{13a^2+5b^2}geqfrac{a+b+c}{18}$ and my second answer I have to prove this :
Let $a,b,c$ be real positive numbers then we have : $$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq frac{ab+bc+ca}{18^2}$$
My try :
With Chebychev's inequality we get :
$$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq left(frac{ab+bc+ca}{3}right)left(sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)right) $$
Remains to prove :
$$sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)geq frac{3}{18^2}$$
Or with the right substitution :
$$sum_{cyc}left(frac{x}{13x+5y}right)left(frac{y}{13y+5z}right)geq frac{3}{18^2}$$
And after this I don't know what to do...
Can someone help me ?
Thanks
real-analysis inequality contest-math
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
asked Nov 23 at 8:01
max8128
1,031421
It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question math.stackexchange.com/q/1777075/42969.
– Martin R
Nov 23 at 8:58
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Related to this Inequality $sumlimits_{cyc}frac{a^3}{13a^2+5b^2}geqfrac{a+b+c}{18}$ and my second answer I have to prove this :
Let $a,b,c$ be real positive numbers then we have : $$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq frac{ab+bc+ca}{18^2}$$
My try :
With Chebychev's inequality we get :
$$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq left(frac{ab+bc+ca}{3}right)left(sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)right) $$
Remains to prove :
$$sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)geq frac{3}{18^2}$$
Or with the right substitution :
$$sum_{cyc}left(frac{x}{13x+5y}right)left(frac{y}{13y+5z}right)geq frac{3}{18^2}$$
And after this I don't know what to do...
Can someone help me ?
Thanks
real-analysis inequality contest-math
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
asked Nov 23 at 8:01
max8128
1,031421
Related to this Inequality $sumlimits_{cyc}frac{a^3}{13a^2+5b^2}geqfrac{a+b+c}{18}$ and my second answer I have to prove this :
Let $a,b,c$ be real positive numbers then we have : $$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq frac{ab+bc+ca}{18^2}$$
My try :
With Chebychev's inequality we get :
$$sum_{cyc}left(frac{a^3}{13a^2+5b^2}right)left(frac{b^3}{13b^2+5c^2}right)geq left(frac{ab+bc+ca}{3}right)left(sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)right) $$
Remains to prove :
$$sum_{cyc}left(frac{a^2}{13a^2+5b^2}right)left(frac{b^2}{13b^2+5c^2}right)geq frac{3}{18^2}$$
Or with the right substitution :
$$sum_{cyc}left(frac{x}{13x+5y}right)left(frac{y}{13y+5z}right)geq frac{3}{18^2}$$
And after this I don't know what to do...
Can someone help me ?
Thanks
real-analysis inequality contest-math
real-analysis inequality contest-math
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
asked Nov 23 at 8:01
max8128
1,031421
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
asked Nov 23 at 8:01
max8128
1,031421
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
edited Nov 23 at 8:07
Chinnapparaj R
4,9871825
4,9871825
asked Nov 23 at 8:01
max8128
1,031421
asked Nov 23 at 8:01
max8128
1,031421
asked Nov 23 at 8:01
max8128
1,031421
1,031421
It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question math.stackexchange.com/q/1777075/42969.
– Martin R
Nov 23 at 8:58
add a comment |
It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question math.stackexchange.com/q/1777075/42969.
– Martin R
Nov 23 at 8:58
It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question math.stackexchange.com/q/1777075/42969.
– Martin R
Nov 23 at 8:58
It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question math.stackexchange.com/q/1777075/42969.
– Martin R
Nov 23 at 8:58
add a comment |
1 Answer
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It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=frac{1}{18cdot33}+frac{8}{33cdot57}+frac{8}{57cdot18}=frac{155}{11286}$$ and
$$RS=frac{5}{324}.$$
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
1
down vote
It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=frac{1}{18cdot33}+frac{8}{33cdot57}+frac{8}{57cdot18}=frac{155}{11286}$$ and
$$RS=frac{5}{324}.$$
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
add a comment |
up vote
1
down vote
It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=frac{1}{18cdot33}+frac{8}{33cdot57}+frac{8}{57cdot18}=frac{155}{11286}$$ and
$$RS=frac{5}{324}.$$
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
add a comment |
up vote
1
down vote
up vote
1
down vote
It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=frac{1}{18cdot33}+frac{8}{33cdot57}+frac{8}{57cdot18}=frac{155}{11286}$$ and
$$RS=frac{5}{324}.$$
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=frac{1}{18cdot33}+frac{8}{33cdot57}+frac{8}{57cdot18}=frac{155}{11286}$$ and
$$RS=frac{5}{324}.$$
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
edited Nov 23 at 8:30
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
answered Nov 23 at 8:23
Michael Rozenberg
94.9k1588183
94.9k1588183
add a comment |
add a comment |
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It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question math.stackexchange.com/q/1777075/42969.
– Martin R
Nov 23 at 8:58