Mixing
Prove $3leqq xsqrt{1+ y^{3}}+ ysqrt{1+ z^{3}}+ zsqrt{1+ x^{3}}leqq 5$
$begingroup$
This is an old problem of Pham Kim Hung!
Prove: $$3leqq xsqrt{1+ y^{3}}+ ysqrt{1+ z^{3}}+ zsqrt{1+ x^{3}}leqq 5$$ with $x,,y,,zgeqq 0$ & $x+y+z=3$
For the LHS, we have:
$$left { sumlimits_{cyc}xsqrt{1+ y^{3}} right }^{2}= sumlimits_{cyc}left { x^{2}+ x^{3}z^{2}+ 2,xysqrt{left ( 1+ y^{3} right )left ( 1+ z^{3} right )} right }geqq sumlimits_{cyc}left { x^{2}+ 2,xy right }= 9$$
And the RHS, we also have:
$$sumlimits_{cyc}xsqrt{1+ y^{3}} = sumlimits_{cyc} xsqrt{left ( 1+ y right )left ( 1+ y^{2}- y right )}leqq sumlimits_{cyc}frac{xleft { 2+ y^{2} right }}{2}= 3+ frac{1}{2}sumlimits_{cyc}xy^{2}leqq 5$$
So, we need to prove: $xy^{2}+ yz^{2}+ zx^{2}leqq 4$ with $x= 3- y- z$
or $$4- xy^{2}- yz^{2}- zx^{2}= left { 3- y- z right }y^{2}- yz^{2}- zleft { 3- y- z right }^{2}= left { y- 1 right }^{2}left { frac{underbrace{y+ 4,z- 5left ( 3- y- z right )}_{y+ 4,z- 5,xgeqq 0}}{3} right }+ left { 3- y- z right }left { frac{left ( 2,y+ 2,z- 3 right )^{2}+ 3}{4} right }geqq 0$$
with $x= minleft { x,,y,,z right }$
Any idea? Who can help me with another solution? I hope to see that! Thanks!
inequality radicals substitution a.m.-g.m.-inequality rearrangement-inequality
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
This is an old problem of Pham Kim Hung!
Prove: $$3leqq xsqrt{1+ y^{3}}+ ysqrt{1+ z^{3}}+ zsqrt{1+ x^{3}}leqq 5$$ with $x,,y,,zgeqq 0$ & $x+y+z=3$
For the LHS, we have:
$$left { sumlimits_{cyc}xsqrt{1+ y^{3}} right }^{2}= sumlimits_{cyc}left { x^{2}+ x^{3}z^{2}+ 2,xysqrt{left ( 1+ y^{3} right )left ( 1+ z^{3} right )} right }geqq sumlimits_{cyc}left { x^{2}+ 2,xy right }= 9$$
And the RHS, we also have:
$$sumlimits_{cyc}xsqrt{1+ y^{3}} = sumlimits_{cyc} xsqrt{left ( 1+ y right )left ( 1+ y^{2}- y right )}leqq sumlimits_{cyc}frac{xleft { 2+ y^{2} right }}{2}= 3+ frac{1}{2}sumlimits_{cyc}xy^{2}leqq 5$$
So, we need to prove: $xy^{2}+ yz^{2}+ zx^{2}leqq 4$ with $x= 3- y- z$
or $$4- xy^{2}- yz^{2}- zx^{2}= left { 3- y- z right }y^{2}- yz^{2}- zleft { 3- y- z right }^{2}= left { y- 1 right }^{2}left { frac{underbrace{y+ 4,z- 5left ( 3- y- z right )}_{y+ 4,z- 5,xgeqq 0}}{3} right }+ left { 3- y- z right }left { frac{left ( 2,y+ 2,z- 3 right )^{2}+ 3}{4} right }geqq 0$$
with $x= minleft { x,,y,,z right }$
Any idea? Who can help me with another solution? I hope to see that! Thanks!
inequality radicals substitution a.m.-g.m.-inequality rearrangement-inequality
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
$endgroup$
$begingroup$
As stated, the problem is incorrect, by taking $x,y,z$ very large, say all equal to $100$. Please correct the title statement.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 2 '18 at 9:17
$begingroup$
I think we need some condition.
$endgroup$
– Michael Rozenberg
Jun 2 '18 at 9:17
$begingroup$
Possible duplicate of Proof of the inequality $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b} geq frac{3}{2}$
$endgroup$
– Chris Custer
Jun 4 '18 at 0:49
add a comment |
$begingroup$
This is an old problem of Pham Kim Hung!
Prove: $$3leqq xsqrt{1+ y^{3}}+ ysqrt{1+ z^{3}}+ zsqrt{1+ x^{3}}leqq 5$$ with $x,,y,,zgeqq 0$ & $x+y+z=3$
For the LHS, we have:
$$left { sumlimits_{cyc}xsqrt{1+ y^{3}} right }^{2}= sumlimits_{cyc}left { x^{2}+ x^{3}z^{2}+ 2,xysqrt{left ( 1+ y^{3} right )left ( 1+ z^{3} right )} right }geqq sumlimits_{cyc}left { x^{2}+ 2,xy right }= 9$$
And the RHS, we also have:
$$sumlimits_{cyc}xsqrt{1+ y^{3}} = sumlimits_{cyc} xsqrt{left ( 1+ y right )left ( 1+ y^{2}- y right )}leqq sumlimits_{cyc}frac{xleft { 2+ y^{2} right }}{2}= 3+ frac{1}{2}sumlimits_{cyc}xy^{2}leqq 5$$
So, we need to prove: $xy^{2}+ yz^{2}+ zx^{2}leqq 4$ with $x= 3- y- z$
or $$4- xy^{2}- yz^{2}- zx^{2}= left { 3- y- z right }y^{2}- yz^{2}- zleft { 3- y- z right }^{2}= left { y- 1 right }^{2}left { frac{underbrace{y+ 4,z- 5left ( 3- y- z right )}_{y+ 4,z- 5,xgeqq 0}}{3} right }+ left { 3- y- z right }left { frac{left ( 2,y+ 2,z- 3 right )^{2}+ 3}{4} right }geqq 0$$
with $x= minleft { x,,y,,z right }$
Any idea? Who can help me with another solution? I hope to see that! Thanks!
inequality radicals substitution a.m.-g.m.-inequality rearrangement-inequality
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
$endgroup$
This is an old problem of Pham Kim Hung!
Prove: $$3leqq xsqrt{1+ y^{3}}+ ysqrt{1+ z^{3}}+ zsqrt{1+ x^{3}}leqq 5$$ with $x,,y,,zgeqq 0$ & $x+y+z=3$
For the LHS, we have:
$$left { sumlimits_{cyc}xsqrt{1+ y^{3}} right }^{2}= sumlimits_{cyc}left { x^{2}+ x^{3}z^{2}+ 2,xysqrt{left ( 1+ y^{3} right )left ( 1+ z^{3} right )} right }geqq sumlimits_{cyc}left { x^{2}+ 2,xy right }= 9$$
And the RHS, we also have:
$$sumlimits_{cyc}xsqrt{1+ y^{3}} = sumlimits_{cyc} xsqrt{left ( 1+ y right )left ( 1+ y^{2}- y right )}leqq sumlimits_{cyc}frac{xleft { 2+ y^{2} right }}{2}= 3+ frac{1}{2}sumlimits_{cyc}xy^{2}leqq 5$$
So, we need to prove: $xy^{2}+ yz^{2}+ zx^{2}leqq 4$ with $x= 3- y- z$
or $$4- xy^{2}- yz^{2}- zx^{2}= left { 3- y- z right }y^{2}- yz^{2}- zleft { 3- y- z right }^{2}= left { y- 1 right }^{2}left { frac{underbrace{y+ 4,z- 5left ( 3- y- z right )}_{y+ 4,z- 5,xgeqq 0}}{3} right }+ left { 3- y- z right }left { frac{left ( 2,y+ 2,z- 3 right )^{2}+ 3}{4} right }geqq 0$$
with $x= minleft { x,,y,,z right }$
Any idea? Who can help me with another solution? I hope to see that! Thanks!
inequality radicals substitution a.m.-g.m.-inequality rearrangement-inequality
inequality radicals substitution a.m.-g.m.-inequality rearrangement-inequality
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
edited Feb 19 at 0:55
Michael Rozenberg
104k1892197
104k1892197
asked Jun 2 '18 at 8:59
user548665
$begingroup$
As stated, the problem is incorrect, by taking $x,y,z$ very large, say all equal to $100$. Please correct the title statement.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 2 '18 at 9:17
$begingroup$
I think we need some condition.
$endgroup$
– Michael Rozenberg
Jun 2 '18 at 9:17
$begingroup$
Possible duplicate of Proof of the inequality $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b} geq frac{3}{2}$
$endgroup$
– Chris Custer
Jun 4 '18 at 0:49
add a comment |
$begingroup$
As stated, the problem is incorrect, by taking $x,y,z$ very large, say all equal to $100$. Please correct the title statement.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 2 '18 at 9:17
$begingroup$
I think we need some condition.
$endgroup$
– Michael Rozenberg
Jun 2 '18 at 9:17
$begingroup$
Possible duplicate of Proof of the inequality $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b} geq frac{3}{2}$
$endgroup$
– Chris Custer
Jun 4 '18 at 0:49
$begingroup$
As stated, the problem is incorrect, by taking $x,y,z$ very large, say all equal to $100$. Please correct the title statement.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 2 '18 at 9:17
$begingroup$
As stated, the problem is incorrect, by taking $x,y,z$ very large, say all equal to $100$. Please correct the title statement.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 2 '18 at 9:17
$begingroup$
I think we need some condition.
$endgroup$
– Michael Rozenberg
Jun 2 '18 at 9:17
$begingroup$
I think we need some condition.
$endgroup$
– Michael Rozenberg
Jun 2 '18 at 9:17
$begingroup$
Possible duplicate of Proof of the inequality $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b} geq frac{3}{2}$
$endgroup$
– Chris Custer
Jun 4 '18 at 0:49
$begingroup$
Possible duplicate of Proof of the inequality $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b} geq frac{3}{2}$
$endgroup$
– Chris Custer
Jun 4 '18 at 0:49
add a comment |
1 Answer
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$begingroup$
I think you mean $x+y+z=3.$
The left inequality:
$$sum_{cyc}xsqrt{1+y^3}geqsum_{cyc}x=3.$$
The right inequality.
Let ${x,y,z}={a,b,c}$, where $ageq bgeq c.$
Thus, by AM-GM, Rearrangement and AM-GM we obtain:
$$sum_{cyc}xsqrt{1+y^3}=sum_{cyc}xsqrt{(1+y)(1-y+y^2)}leqfrac{1}{2}sum_{cyc}x(1+y+1-y+y^2)=$$
$$=3+frac{1}{2}(xy^2+yz^2+zx^2)=3+frac{1}{2}(xycdot y+yzcdot z+zxcdot x)leq$$
$$leq3+frac{1}{2}(abcdot a+accdot b+bccdot c)=3+frac{1}{2}b(a^2+ac+c^2)leq$$
$$leq3+frac{1}{4}cdot2b(a+c)^2leq3+frac{1}{4}left(frac{2b+a+c+a+c}{3}right)^3=5.$$
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
add a comment |
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1 Answer
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$begingroup$
I think you mean $x+y+z=3.$
The left inequality:
$$sum_{cyc}xsqrt{1+y^3}geqsum_{cyc}x=3.$$
The right inequality.
Let ${x,y,z}={a,b,c}$, where $ageq bgeq c.$
Thus, by AM-GM, Rearrangement and AM-GM we obtain:
$$sum_{cyc}xsqrt{1+y^3}=sum_{cyc}xsqrt{(1+y)(1-y+y^2)}leqfrac{1}{2}sum_{cyc}x(1+y+1-y+y^2)=$$
$$=3+frac{1}{2}(xy^2+yz^2+zx^2)=3+frac{1}{2}(xycdot y+yzcdot z+zxcdot x)leq$$
$$leq3+frac{1}{2}(abcdot a+accdot b+bccdot c)=3+frac{1}{2}b(a^2+ac+c^2)leq$$
$$leq3+frac{1}{4}cdot2b(a+c)^2leq3+frac{1}{4}left(frac{2b+a+c+a+c}{3}right)^3=5.$$
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
add a comment |
$begingroup$
I think you mean $x+y+z=3.$
The left inequality:
$$sum_{cyc}xsqrt{1+y^3}geqsum_{cyc}x=3.$$
The right inequality.
Let ${x,y,z}={a,b,c}$, where $ageq bgeq c.$
Thus, by AM-GM, Rearrangement and AM-GM we obtain:
$$sum_{cyc}xsqrt{1+y^3}=sum_{cyc}xsqrt{(1+y)(1-y+y^2)}leqfrac{1}{2}sum_{cyc}x(1+y+1-y+y^2)=$$
$$=3+frac{1}{2}(xy^2+yz^2+zx^2)=3+frac{1}{2}(xycdot y+yzcdot z+zxcdot x)leq$$
$$leq3+frac{1}{2}(abcdot a+accdot b+bccdot c)=3+frac{1}{2}b(a^2+ac+c^2)leq$$
$$leq3+frac{1}{4}cdot2b(a+c)^2leq3+frac{1}{4}left(frac{2b+a+c+a+c}{3}right)^3=5.$$
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
add a comment |
$begingroup$
I think you mean $x+y+z=3.$
The left inequality:
$$sum_{cyc}xsqrt{1+y^3}geqsum_{cyc}x=3.$$
The right inequality.
Let ${x,y,z}={a,b,c}$, where $ageq bgeq c.$
Thus, by AM-GM, Rearrangement and AM-GM we obtain:
$$sum_{cyc}xsqrt{1+y^3}=sum_{cyc}xsqrt{(1+y)(1-y+y^2)}leqfrac{1}{2}sum_{cyc}x(1+y+1-y+y^2)=$$
$$=3+frac{1}{2}(xy^2+yz^2+zx^2)=3+frac{1}{2}(xycdot y+yzcdot z+zxcdot x)leq$$
$$leq3+frac{1}{2}(abcdot a+accdot b+bccdot c)=3+frac{1}{2}b(a^2+ac+c^2)leq$$
$$leq3+frac{1}{4}cdot2b(a+c)^2leq3+frac{1}{4}left(frac{2b+a+c+a+c}{3}right)^3=5.$$
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
I think you mean $x+y+z=3.$
The left inequality:
$$sum_{cyc}xsqrt{1+y^3}geqsum_{cyc}x=3.$$
The right inequality.
Let ${x,y,z}={a,b,c}$, where $ageq bgeq c.$
Thus, by AM-GM, Rearrangement and AM-GM we obtain:
$$sum_{cyc}xsqrt{1+y^3}=sum_{cyc}xsqrt{(1+y)(1-y+y^2)}leqfrac{1}{2}sum_{cyc}x(1+y+1-y+y^2)=$$
$$=3+frac{1}{2}(xy^2+yz^2+zx^2)=3+frac{1}{2}(xycdot y+yzcdot z+zxcdot x)leq$$
$$leq3+frac{1}{2}(abcdot a+accdot b+bccdot c)=3+frac{1}{2}b(a^2+ac+c^2)leq$$
$$leq3+frac{1}{4}cdot2b(a+c)^2leq3+frac{1}{4}left(frac{2b+a+c+a+c}{3}right)^3=5.$$
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
edited Jun 2 '18 at 9:30
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
answered Jun 2 '18 at 9:20
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
add a comment |
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
$begingroup$
@Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted?
$endgroup$
– Michael Rozenberg
Feb 17 at 6:51
add a comment |
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$begingroup$
As stated, the problem is incorrect, by taking $x,y,z$ very large, say all equal to $100$. Please correct the title statement.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 2 '18 at 9:17
$begingroup$
I think we need some condition.
$endgroup$
– Michael Rozenberg
Jun 2 '18 at 9:17
$begingroup$
Possible duplicate of Proof of the inequality $frac{a}{b+c}+frac{b}{a+c}+frac{c}{a+b} geq frac{3}{2}$
$endgroup$
– Chris Custer
Jun 4 '18 at 0:49