Mixing
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prove $sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}leq 1$
If $x$,$y$,$z$ are positive real numbers,Prove:$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}leq 1$$
Using this two inequality:
$sum ^n_{i=1} sqrt{a_ib_i}leqsqrt {ab} $ (we call it $A$ inequality)
$frac {ab}{a+b} geq sum ^n_{i=1} frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)
which $a_i$ and $b_i$ are positive and and $b= sum ^n_{i=1} b_i$,$a= sum ^n_{i=1} a_i$.
Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.
Things I have tried so far:
Using inequality $A$ i can re write question inequality as$$sum limits_{cyc} frac{x}{2x+sqrt{xy}}leq 1$$
And I can't go further.I can't observer something that could lead me to using inequality $B$.
inequality
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 30 '14 at 13:57
user2838619
1,6231128
add a comment |
If $x$,$y$,$z$ are positive real numbers,Prove:$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}leq 1$$
Using this two inequality:
$sum ^n_{i=1} sqrt{a_ib_i}leqsqrt {ab} $ (we call it $A$ inequality)
$frac {ab}{a+b} geq sum ^n_{i=1} frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)
which $a_i$ and $b_i$ are positive and and $b= sum ^n_{i=1} b_i$,$a= sum ^n_{i=1} a_i$.
Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.
Things I have tried so far:
Using inequality $A$ i can re write question inequality as$$sum limits_{cyc} frac{x}{2x+sqrt{xy}}leq 1$$
And I can't go further.I can't observer something that could lead me to using inequality $B$.
inequality
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 30 '14 at 13:57
user2838619
1,6231128
add a comment |
If $x$,$y$,$z$ are positive real numbers,Prove:$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}leq 1$$
Using this two inequality:
$sum ^n_{i=1} sqrt{a_ib_i}leqsqrt {ab} $ (we call it $A$ inequality)
$frac {ab}{a+b} geq sum ^n_{i=1} frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)
which $a_i$ and $b_i$ are positive and and $b= sum ^n_{i=1} b_i$,$a= sum ^n_{i=1} a_i$.
Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.
Things I have tried so far:
Using inequality $A$ i can re write question inequality as$$sum limits_{cyc} frac{x}{2x+sqrt{xy}}leq 1$$
And I can't go further.I can't observer something that could lead me to using inequality $B$.
inequality
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 30 '14 at 13:57
user2838619
1,6231128
If $x$,$y$,$z$ are positive real numbers,Prove:$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}leq 1$$
Using this two inequality:
$sum ^n_{i=1} sqrt{a_ib_i}leqsqrt {ab} $ (we call it $A$ inequality)
$frac {ab}{a+b} geq sum ^n_{i=1} frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)
which $a_i$ and $b_i$ are positive and and $b= sum ^n_{i=1} b_i$,$a= sum ^n_{i=1} a_i$.
Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.
Things I have tried so far:
Using inequality $A$ i can re write question inequality as$$sum limits_{cyc} frac{x}{2x+sqrt{xy}}leq 1$$
And I can't go further.I can't observer something that could lead me to using inequality $B$.
inequality
inequality
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 30 '14 at 13:57
user2838619
1,6231128
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 30 '14 at 13:57
user2838619
1,6231128
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Jul 30 '14 at 13:57
user2838619
1,6231128
asked Jul 30 '14 at 13:57
user2838619
1,6231128
asked Jul 30 '14 at 13:57
user2838619
1,6231128
1,6231128
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Use the fact
$sqrt{(x+y)(x+z)}ge sqrt{xy}+sqrt{xz}$. Which is obvious after squaring and cancelling terms.
Now our inequality becomes :
$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}le sum limits_{cyc} frac{x}{x+sqrt{xy}+sqrt{xz}}=sum limits_{cyc} dfrac{sqrt{x}}{sqrt{x}+sqrt{y}+sqrt{z}}=1$$
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
answered Jul 30 '14 at 14:15
shadow10
2,855931
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
2
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
add a comment |
By your work and by C-S we obtain:
$$sum_{cyc}frac{x}{x+sqrt{(x+y)(x+z)}}leqsum_{cyc}frac{x}{2x+sqrt{yz}}=frac{3}{2}-sum_{cyc}left(frac{1}{2}-frac{x}{2x+sqrt{yz}}right)=$$
$$=frac{3}{2}-frac{1}{2}sum_{cyc}frac{sqrt{yz}}{2x+sqrt{yz}}=frac{3}{2}-frac{1}{2}sum_{cyc}frac{yz}{2xsqrt{yz}+yz}leq$$
$$leqfrac{3}{2}-frac{1}{2}cdotfrac{left(sumlimits_{cyc}sqrt{yz}right)^2}{sumlimits_{cyc}(2xsqrt{yz}+yz)}=frac{3}{2}-frac{1}{2}=1.$$
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
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Use the fact
$sqrt{(x+y)(x+z)}ge sqrt{xy}+sqrt{xz}$. Which is obvious after squaring and cancelling terms.
Now our inequality becomes :
$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}le sum limits_{cyc} frac{x}{x+sqrt{xy}+sqrt{xz}}=sum limits_{cyc} dfrac{sqrt{x}}{sqrt{x}+sqrt{y}+sqrt{z}}=1$$
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
answered Jul 30 '14 at 14:15
shadow10
2,855931
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
2
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
add a comment |
Use the fact
$sqrt{(x+y)(x+z)}ge sqrt{xy}+sqrt{xz}$. Which is obvious after squaring and cancelling terms.
Now our inequality becomes :
$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}le sum limits_{cyc} frac{x}{x+sqrt{xy}+sqrt{xz}}=sum limits_{cyc} dfrac{sqrt{x}}{sqrt{x}+sqrt{y}+sqrt{z}}=1$$
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
answered Jul 30 '14 at 14:15
shadow10
2,855931
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
2
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
add a comment |
Use the fact
$sqrt{(x+y)(x+z)}ge sqrt{xy}+sqrt{xz}$. Which is obvious after squaring and cancelling terms.
Now our inequality becomes :
$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}le sum limits_{cyc} frac{x}{x+sqrt{xy}+sqrt{xz}}=sum limits_{cyc} dfrac{sqrt{x}}{sqrt{x}+sqrt{y}+sqrt{z}}=1$$
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
answered Jul 30 '14 at 14:15
shadow10
2,855931
Use the fact
$sqrt{(x+y)(x+z)}ge sqrt{xy}+sqrt{xz}$. Which is obvious after squaring and cancelling terms.
Now our inequality becomes :
$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}le sum limits_{cyc} frac{x}{x+sqrt{xy}+sqrt{xz}}=sum limits_{cyc} dfrac{sqrt{x}}{sqrt{x}+sqrt{y}+sqrt{z}}=1$$
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
answered Jul 30 '14 at 14:15
shadow10
2,855931
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
edited Jan 28 '18 at 14:07
Vee Hua Zhi
759224
759224
answered Jul 30 '14 at 14:15
shadow10
2,855931
answered Jul 30 '14 at 14:15
shadow10
2,855931
answered Jul 30 '14 at 14:15
shadow10
2,855931
2,855931
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
2
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
add a comment |
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
2
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
thanks for hint,i'm trying figuring out the part you write $sum frac{x}{x+sqrt{xy}+sqrt{xz}}=1$.can you give me a hint on how you concluded that?
– user2838619
Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea.
– shadow10
Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
How i missed a simple factoring like that.thanks.
– user2838619
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
So only inequality A is needed.
– Ewan Delanoy
Jul 30 '14 at 15:19
2
2
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
funny thing is Mathematica has serious problem with it
– user2838619
Jul 30 '14 at 16:19
add a comment |
By your work and by C-S we obtain:
$$sum_{cyc}frac{x}{x+sqrt{(x+y)(x+z)}}leqsum_{cyc}frac{x}{2x+sqrt{yz}}=frac{3}{2}-sum_{cyc}left(frac{1}{2}-frac{x}{2x+sqrt{yz}}right)=$$
$$=frac{3}{2}-frac{1}{2}sum_{cyc}frac{sqrt{yz}}{2x+sqrt{yz}}=frac{3}{2}-frac{1}{2}sum_{cyc}frac{yz}{2xsqrt{yz}+yz}leq$$
$$leqfrac{3}{2}-frac{1}{2}cdotfrac{left(sumlimits_{cyc}sqrt{yz}right)^2}{sumlimits_{cyc}(2xsqrt{yz}+yz)}=frac{3}{2}-frac{1}{2}=1.$$
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
add a comment |
By your work and by C-S we obtain:
$$sum_{cyc}frac{x}{x+sqrt{(x+y)(x+z)}}leqsum_{cyc}frac{x}{2x+sqrt{yz}}=frac{3}{2}-sum_{cyc}left(frac{1}{2}-frac{x}{2x+sqrt{yz}}right)=$$
$$=frac{3}{2}-frac{1}{2}sum_{cyc}frac{sqrt{yz}}{2x+sqrt{yz}}=frac{3}{2}-frac{1}{2}sum_{cyc}frac{yz}{2xsqrt{yz}+yz}leq$$
$$leqfrac{3}{2}-frac{1}{2}cdotfrac{left(sumlimits_{cyc}sqrt{yz}right)^2}{sumlimits_{cyc}(2xsqrt{yz}+yz)}=frac{3}{2}-frac{1}{2}=1.$$
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
add a comment |
By your work and by C-S we obtain:
$$sum_{cyc}frac{x}{x+sqrt{(x+y)(x+z)}}leqsum_{cyc}frac{x}{2x+sqrt{yz}}=frac{3}{2}-sum_{cyc}left(frac{1}{2}-frac{x}{2x+sqrt{yz}}right)=$$
$$=frac{3}{2}-frac{1}{2}sum_{cyc}frac{sqrt{yz}}{2x+sqrt{yz}}=frac{3}{2}-frac{1}{2}sum_{cyc}frac{yz}{2xsqrt{yz}+yz}leq$$
$$leqfrac{3}{2}-frac{1}{2}cdotfrac{left(sumlimits_{cyc}sqrt{yz}right)^2}{sumlimits_{cyc}(2xsqrt{yz}+yz)}=frac{3}{2}-frac{1}{2}=1.$$
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
By your work and by C-S we obtain:
$$sum_{cyc}frac{x}{x+sqrt{(x+y)(x+z)}}leqsum_{cyc}frac{x}{2x+sqrt{yz}}=frac{3}{2}-sum_{cyc}left(frac{1}{2}-frac{x}{2x+sqrt{yz}}right)=$$
$$=frac{3}{2}-frac{1}{2}sum_{cyc}frac{sqrt{yz}}{2x+sqrt{yz}}=frac{3}{2}-frac{1}{2}sum_{cyc}frac{yz}{2xsqrt{yz}+yz}leq$$
$$leqfrac{3}{2}-frac{1}{2}cdotfrac{left(sumlimits_{cyc}sqrt{yz}right)^2}{sumlimits_{cyc}(2xsqrt{yz}+yz)}=frac{3}{2}-frac{1}{2}=1.$$
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
answered Nov 21 '18 at 3:07
Michael Rozenberg
96.8k1589188
96.8k1589188
add a comment |
add a comment |
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