Mixing
Inequality triangle Radon substitutions
$begingroup$
I have this inequality:
$$sum frac {a^3}{p-a}geq 8(2R-r)^2$$
I have tried using Radon substitutions and I get this:
$$sum frac{(y+x)^3}{x}geq 8(2R-r)^2$$
I know from Holder that :
$$sum frac{(y+x)^3}{x}geq frac{(2(x+y+z))^3}{3(x+y+z)}=frac{8(x+y+z)^2}{3}$$
Now I want to prove that:
$$frac{8(x+y+z)^2}{3}geq 8(2R-r)^2$$
And after that I get:
$$p^2geq 48(2R-r)^2$$
Now here I'm not sure that I can prove this.
If someone has a hint or some help I would apreciate.
inequality substitution cauchy-schwarz-inequality geometric-inequalities sos
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
asked Jan 15 at 13:27
ovetz13ovetz13
1547
$endgroup$
add a comment |
$begingroup$
I have this inequality:
$$sum frac {a^3}{p-a}geq 8(2R-r)^2$$
I have tried using Radon substitutions and I get this:
$$sum frac{(y+x)^3}{x}geq 8(2R-r)^2$$
I know from Holder that :
$$sum frac{(y+x)^3}{x}geq frac{(2(x+y+z))^3}{3(x+y+z)}=frac{8(x+y+z)^2}{3}$$
Now I want to prove that:
$$frac{8(x+y+z)^2}{3}geq 8(2R-r)^2$$
And after that I get:
$$p^2geq 48(2R-r)^2$$
Now here I'm not sure that I can prove this.
If someone has a hint or some help I would apreciate.
inequality substitution cauchy-schwarz-inequality geometric-inequalities sos
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
asked Jan 15 at 13:27
ovetz13ovetz13
1547
$endgroup$
add a comment |
$begingroup$
I have this inequality:
$$sum frac {a^3}{p-a}geq 8(2R-r)^2$$
I have tried using Radon substitutions and I get this:
$$sum frac{(y+x)^3}{x}geq 8(2R-r)^2$$
I know from Holder that :
$$sum frac{(y+x)^3}{x}geq frac{(2(x+y+z))^3}{3(x+y+z)}=frac{8(x+y+z)^2}{3}$$
Now I want to prove that:
$$frac{8(x+y+z)^2}{3}geq 8(2R-r)^2$$
And after that I get:
$$p^2geq 48(2R-r)^2$$
Now here I'm not sure that I can prove this.
If someone has a hint or some help I would apreciate.
inequality substitution cauchy-schwarz-inequality geometric-inequalities sos
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
asked Jan 15 at 13:27
ovetz13ovetz13
1547
$endgroup$
I have this inequality:
$$sum frac {a^3}{p-a}geq 8(2R-r)^2$$
I have tried using Radon substitutions and I get this:
$$sum frac{(y+x)^3}{x}geq 8(2R-r)^2$$
I know from Holder that :
$$sum frac{(y+x)^3}{x}geq frac{(2(x+y+z))^3}{3(x+y+z)}=frac{8(x+y+z)^2}{3}$$
Now I want to prove that:
$$frac{8(x+y+z)^2}{3}geq 8(2R-r)^2$$
And after that I get:
$$p^2geq 48(2R-r)^2$$
Now here I'm not sure that I can prove this.
If someone has a hint or some help I would apreciate.
inequality substitution cauchy-schwarz-inequality geometric-inequalities sos
inequality substitution cauchy-schwarz-inequality geometric-inequalities sos
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
asked Jan 15 at 13:27
ovetz13ovetz13
1547
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
asked Jan 15 at 13:27
ovetz13ovetz13
1547
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
edited Jan 15 at 21:08
Michael Rozenberg
104k1891196
104k1891196
asked Jan 15 at 13:27
ovetz13ovetz13
1547
asked Jan 15 at 13:27
ovetz13ovetz13
1547
asked Jan 15 at 13:27
ovetz13ovetz13
1547
1547
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the standard notation we need to prove that
$$sum_{cyc}frac{a^3}{b+c-a}geq4left(frac{abc}{2S}-frac{2S}{a+b+c}right)^2.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(abc(a+b+c)-4S^2)^2}{S^2(a+b+c)^2}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(2(x+y)(x+z)(y+z)(x+y+z)-4xyz(x+y+z))^2}{4xyz(x+y+z)^3}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{((x+y)(x+z)(y+z)-2xyz)^2}{xyz(x+y+z)}$$ or
$$(x+y+z)sum_{cyc}xy(x+y)^3geq2left(sum_{cyc}(x^2y+x^2z)right)^2.$$
Now, by C-S
$$(x+y+z)sum_{cyc}xy(x+y)^3=frac{1}{2}sum_{cyc}(x+y)sum_{cyc}xy(x+y)^3geqfrac{1}{2}left(sum_{cyc}sqrt{xy}(x+y)^2right)^2.$$
Thus, it's enough to prove that
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}(x^2y+xy^2)$$ or
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}xy(x+y)$$ or
$$sum_{cyc}sqrt{xy}(x+y)(sqrt{x}-sqrt{y})^2geq0.$$
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
1
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
add a comment |
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$begingroup$
In the standard notation we need to prove that
$$sum_{cyc}frac{a^3}{b+c-a}geq4left(frac{abc}{2S}-frac{2S}{a+b+c}right)^2.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(abc(a+b+c)-4S^2)^2}{S^2(a+b+c)^2}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(2(x+y)(x+z)(y+z)(x+y+z)-4xyz(x+y+z))^2}{4xyz(x+y+z)^3}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{((x+y)(x+z)(y+z)-2xyz)^2}{xyz(x+y+z)}$$ or
$$(x+y+z)sum_{cyc}xy(x+y)^3geq2left(sum_{cyc}(x^2y+x^2z)right)^2.$$
Now, by C-S
$$(x+y+z)sum_{cyc}xy(x+y)^3=frac{1}{2}sum_{cyc}(x+y)sum_{cyc}xy(x+y)^3geqfrac{1}{2}left(sum_{cyc}sqrt{xy}(x+y)^2right)^2.$$
Thus, it's enough to prove that
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}(x^2y+xy^2)$$ or
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}xy(x+y)$$ or
$$sum_{cyc}sqrt{xy}(x+y)(sqrt{x}-sqrt{y})^2geq0.$$
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
1
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
add a comment |
$begingroup$
In the standard notation we need to prove that
$$sum_{cyc}frac{a^3}{b+c-a}geq4left(frac{abc}{2S}-frac{2S}{a+b+c}right)^2.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(abc(a+b+c)-4S^2)^2}{S^2(a+b+c)^2}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(2(x+y)(x+z)(y+z)(x+y+z)-4xyz(x+y+z))^2}{4xyz(x+y+z)^3}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{((x+y)(x+z)(y+z)-2xyz)^2}{xyz(x+y+z)}$$ or
$$(x+y+z)sum_{cyc}xy(x+y)^3geq2left(sum_{cyc}(x^2y+x^2z)right)^2.$$
Now, by C-S
$$(x+y+z)sum_{cyc}xy(x+y)^3=frac{1}{2}sum_{cyc}(x+y)sum_{cyc}xy(x+y)^3geqfrac{1}{2}left(sum_{cyc}sqrt{xy}(x+y)^2right)^2.$$
Thus, it's enough to prove that
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}(x^2y+xy^2)$$ or
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}xy(x+y)$$ or
$$sum_{cyc}sqrt{xy}(x+y)(sqrt{x}-sqrt{y})^2geq0.$$
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
1
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
add a comment |
$begingroup$
In the standard notation we need to prove that
$$sum_{cyc}frac{a^3}{b+c-a}geq4left(frac{abc}{2S}-frac{2S}{a+b+c}right)^2.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(abc(a+b+c)-4S^2)^2}{S^2(a+b+c)^2}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(2(x+y)(x+z)(y+z)(x+y+z)-4xyz(x+y+z))^2}{4xyz(x+y+z)^3}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{((x+y)(x+z)(y+z)-2xyz)^2}{xyz(x+y+z)}$$ or
$$(x+y+z)sum_{cyc}xy(x+y)^3geq2left(sum_{cyc}(x^2y+x^2z)right)^2.$$
Now, by C-S
$$(x+y+z)sum_{cyc}xy(x+y)^3=frac{1}{2}sum_{cyc}(x+y)sum_{cyc}xy(x+y)^3geqfrac{1}{2}left(sum_{cyc}sqrt{xy}(x+y)^2right)^2.$$
Thus, it's enough to prove that
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}(x^2y+xy^2)$$ or
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}xy(x+y)$$ or
$$sum_{cyc}sqrt{xy}(x+y)(sqrt{x}-sqrt{y})^2geq0.$$
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
In the standard notation we need to prove that
$$sum_{cyc}frac{a^3}{b+c-a}geq4left(frac{abc}{2S}-frac{2S}{a+b+c}right)^2.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(abc(a+b+c)-4S^2)^2}{S^2(a+b+c)^2}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{(2(x+y)(x+z)(y+z)(x+y+z)-4xyz(x+y+z))^2}{4xyz(x+y+z)^3}$$ or
$$sum_{cyc}frac{(y+z)^3}{2x}geqfrac{((x+y)(x+z)(y+z)-2xyz)^2}{xyz(x+y+z)}$$ or
$$(x+y+z)sum_{cyc}xy(x+y)^3geq2left(sum_{cyc}(x^2y+x^2z)right)^2.$$
Now, by C-S
$$(x+y+z)sum_{cyc}xy(x+y)^3=frac{1}{2}sum_{cyc}(x+y)sum_{cyc}xy(x+y)^3geqfrac{1}{2}left(sum_{cyc}sqrt{xy}(x+y)^2right)^2.$$
Thus, it's enough to prove that
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}(x^2y+xy^2)$$ or
$$sum_{cyc}sqrt{xy}(x+y)^2geq2sum_{cyc}xy(x+y)$$ or
$$sum_{cyc}sqrt{xy}(x+y)(sqrt{x}-sqrt{y})^2geq0.$$
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
edited Jan 16 at 9:21
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 21:07
Michael RozenbergMichael Rozenberg
104k1891196
104k1891196
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
1
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
add a comment |
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
1
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
You are awesome as always! Have you seen this ?math.stackexchange.com/a/3062532/522272
$endgroup$
– ovetz13
Jan 15 at 21:24
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
From here$sum_{cyc} sqrt {xy}(x+y)^2geq sum_{cyc} (x^2y+x^2z)$ I think we can prove that $$sqrt {xy}(x+y)^2 geq 2(x^2y+y^2x)$$
$endgroup$
– ovetz13
Jan 16 at 9:05
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
$begingroup$
@ovetz13 Yes! It's much more better. I fixed. Thank you!
$endgroup$
– Michael Rozenberg
Jan 16 at 9:18
1
1
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
$begingroup$
Thank you ! I am still learning and I'm happy that I can learn from you.
$endgroup$
– ovetz13
Jan 16 at 9:59
add a comment |
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