Mixing
Inequality problem: with $x,y,z>0$, show that $frac{x^5}{y^3}+frac{y^5}{z^3}+frac{z^5}{x^3}geq...
$begingroup$
I am studying AM-GM inequalities in school and have this problem:
With $x,y,z>0$ show that $frac{x^5}{y^3}+frac{y^5}{z^3}+frac{z^5}{x^3}geq x^2+y^2+z^2$.
inequality a.m.-g.m.-inequality
edited Jan 12 at 2:59
David G. Stork
11k41432
asked Jan 12 at 2:57
HoàngHoàng
82
$endgroup$
add a comment |
$begingroup$
I am studying AM-GM inequalities in school and have this problem:
With $x,y,z>0$ show that $frac{x^5}{y^3}+frac{y^5}{z^3}+frac{z^5}{x^3}geq x^2+y^2+z^2$.
inequality a.m.-g.m.-inequality
edited Jan 12 at 2:59
David G. Stork
11k41432
asked Jan 12 at 2:57
HoàngHoàng
82
$endgroup$
add a comment |
$begingroup$
I am studying AM-GM inequalities in school and have this problem:
With $x,y,z>0$ show that $frac{x^5}{y^3}+frac{y^5}{z^3}+frac{z^5}{x^3}geq x^2+y^2+z^2$.
inequality a.m.-g.m.-inequality
edited Jan 12 at 2:59
David G. Stork
11k41432
asked Jan 12 at 2:57
HoàngHoàng
82
$endgroup$
I am studying AM-GM inequalities in school and have this problem:
With $x,y,z>0$ show that $frac{x^5}{y^3}+frac{y^5}{z^3}+frac{z^5}{x^3}geq x^2+y^2+z^2$.
inequality a.m.-g.m.-inequality
inequality a.m.-g.m.-inequality
edited Jan 12 at 2:59
David G. Stork
11k41432
asked Jan 12 at 2:57
HoàngHoàng
82
edited Jan 12 at 2:59
David G. Stork
11k41432
asked Jan 12 at 2:57
HoàngHoàng
82
edited Jan 12 at 2:59
David G. Stork
11k41432
edited Jan 12 at 2:59
David G. Stork
11k41432
edited Jan 12 at 2:59
David G. Stork
11k41432
11k41432
asked Jan 12 at 2:57
HoàngHoàng
82
asked Jan 12 at 2:57
HoàngHoàng
82
asked Jan 12 at 2:57
HoàngHoàng
82
82
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Idea behind the solution
By AM-GM you have for each $alpha, beta, gamma in mathbb N$
$$frac{alphafrac{x^5}{y^3}+betafrac{y^5}{z^3}+gammafrac{z^5}{x^3}}{alpha+beta+gamma}geq sqrt[alpha+beta+gamma]{(frac{x^5}{y^3})^alpha(frac{y^5}{z^3})^beta(frac{z^5}{x^3})^gamma}$$
Then by symmetry you can permute circularly the coefficients and add the inequalities.
Now, set the RHS $=x^2$. This yields
$$5 alpha -3 gamma = 2alpha+2beta+2gamma \
5 beta- 3 alpha=0 \
5 gamma -3 beta =0
$$
Since the third equation is the sum of the first two (this comes from the fact that your inequality is homogeneous) your system has infinitely many solutions. Solving you get
$$9 alpha =15 beta =25 gamma$$
In particular the following is a solution:
$$alpha=25 \
beta =15 \
gamma=9$$
The solution
By AM-GM you have
$$frac{25frac{x^5}{y^3}+15frac{y^5}{z^3}+9frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{25}(frac{y^5}{z^3})^{15}(frac{z^5}{x^3})^{9}}=x^2\
frac{9frac{x^5}{y^3}+25frac{y^5}{z^3}+15frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{9}(frac{y^5}{z^3})^{25}(frac{z^5}{x^3})^{15}}=y^2\
frac{15frac{x^5}{y^3}+9frac{y^5}{z^3}+25frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{15}(frac{y^5}{z^3})^{9}(frac{z^5}{x^3})^{25}}=z^2\
$$
Add them together.
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
$endgroup$
1
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
add a comment |
$begingroup$
By AM-GM $$2x^5+3y^5geq5sqrt[5]{(x^5)^2(y^5)^3}=5x^2y^3,$$ which gives
$$frac{x^5}{y^3}geqfrac{5x^2-3y^2}{2}.$$
Id est, $$sum_{cyc}frac{x^5}{y^3}geqsum_{cyc}frac{5x^2-3y^2}{2}=sum_{cyc}x^2.$$
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
2
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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$begingroup$
Idea behind the solution
By AM-GM you have for each $alpha, beta, gamma in mathbb N$
$$frac{alphafrac{x^5}{y^3}+betafrac{y^5}{z^3}+gammafrac{z^5}{x^3}}{alpha+beta+gamma}geq sqrt[alpha+beta+gamma]{(frac{x^5}{y^3})^alpha(frac{y^5}{z^3})^beta(frac{z^5}{x^3})^gamma}$$
Then by symmetry you can permute circularly the coefficients and add the inequalities.
Now, set the RHS $=x^2$. This yields
$$5 alpha -3 gamma = 2alpha+2beta+2gamma \
5 beta- 3 alpha=0 \
5 gamma -3 beta =0
$$
Since the third equation is the sum of the first two (this comes from the fact that your inequality is homogeneous) your system has infinitely many solutions. Solving you get
$$9 alpha =15 beta =25 gamma$$
In particular the following is a solution:
$$alpha=25 \
beta =15 \
gamma=9$$
The solution
By AM-GM you have
$$frac{25frac{x^5}{y^3}+15frac{y^5}{z^3}+9frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{25}(frac{y^5}{z^3})^{15}(frac{z^5}{x^3})^{9}}=x^2\
frac{9frac{x^5}{y^3}+25frac{y^5}{z^3}+15frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{9}(frac{y^5}{z^3})^{25}(frac{z^5}{x^3})^{15}}=y^2\
frac{15frac{x^5}{y^3}+9frac{y^5}{z^3}+25frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{15}(frac{y^5}{z^3})^{9}(frac{z^5}{x^3})^{25}}=z^2\
$$
Add them together.
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
$endgroup$
1
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
add a comment |
$begingroup$
Idea behind the solution
By AM-GM you have for each $alpha, beta, gamma in mathbb N$
$$frac{alphafrac{x^5}{y^3}+betafrac{y^5}{z^3}+gammafrac{z^5}{x^3}}{alpha+beta+gamma}geq sqrt[alpha+beta+gamma]{(frac{x^5}{y^3})^alpha(frac{y^5}{z^3})^beta(frac{z^5}{x^3})^gamma}$$
Then by symmetry you can permute circularly the coefficients and add the inequalities.
Now, set the RHS $=x^2$. This yields
$$5 alpha -3 gamma = 2alpha+2beta+2gamma \
5 beta- 3 alpha=0 \
5 gamma -3 beta =0
$$
Since the third equation is the sum of the first two (this comes from the fact that your inequality is homogeneous) your system has infinitely many solutions. Solving you get
$$9 alpha =15 beta =25 gamma$$
In particular the following is a solution:
$$alpha=25 \
beta =15 \
gamma=9$$
The solution
By AM-GM you have
$$frac{25frac{x^5}{y^3}+15frac{y^5}{z^3}+9frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{25}(frac{y^5}{z^3})^{15}(frac{z^5}{x^3})^{9}}=x^2\
frac{9frac{x^5}{y^3}+25frac{y^5}{z^3}+15frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{9}(frac{y^5}{z^3})^{25}(frac{z^5}{x^3})^{15}}=y^2\
frac{15frac{x^5}{y^3}+9frac{y^5}{z^3}+25frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{15}(frac{y^5}{z^3})^{9}(frac{z^5}{x^3})^{25}}=z^2\
$$
Add them together.
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
$endgroup$
1
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
add a comment |
$begingroup$
Idea behind the solution
By AM-GM you have for each $alpha, beta, gamma in mathbb N$
$$frac{alphafrac{x^5}{y^3}+betafrac{y^5}{z^3}+gammafrac{z^5}{x^3}}{alpha+beta+gamma}geq sqrt[alpha+beta+gamma]{(frac{x^5}{y^3})^alpha(frac{y^5}{z^3})^beta(frac{z^5}{x^3})^gamma}$$
Then by symmetry you can permute circularly the coefficients and add the inequalities.
Now, set the RHS $=x^2$. This yields
$$5 alpha -3 gamma = 2alpha+2beta+2gamma \
5 beta- 3 alpha=0 \
5 gamma -3 beta =0
$$
Since the third equation is the sum of the first two (this comes from the fact that your inequality is homogeneous) your system has infinitely many solutions. Solving you get
$$9 alpha =15 beta =25 gamma$$
In particular the following is a solution:
$$alpha=25 \
beta =15 \
gamma=9$$
The solution
By AM-GM you have
$$frac{25frac{x^5}{y^3}+15frac{y^5}{z^3}+9frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{25}(frac{y^5}{z^3})^{15}(frac{z^5}{x^3})^{9}}=x^2\
frac{9frac{x^5}{y^3}+25frac{y^5}{z^3}+15frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{9}(frac{y^5}{z^3})^{25}(frac{z^5}{x^3})^{15}}=y^2\
frac{15frac{x^5}{y^3}+9frac{y^5}{z^3}+25frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{15}(frac{y^5}{z^3})^{9}(frac{z^5}{x^3})^{25}}=z^2\
$$
Add them together.
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
$endgroup$
Idea behind the solution
By AM-GM you have for each $alpha, beta, gamma in mathbb N$
$$frac{alphafrac{x^5}{y^3}+betafrac{y^5}{z^3}+gammafrac{z^5}{x^3}}{alpha+beta+gamma}geq sqrt[alpha+beta+gamma]{(frac{x^5}{y^3})^alpha(frac{y^5}{z^3})^beta(frac{z^5}{x^3})^gamma}$$
Then by symmetry you can permute circularly the coefficients and add the inequalities.
Now, set the RHS $=x^2$. This yields
$$5 alpha -3 gamma = 2alpha+2beta+2gamma \
5 beta- 3 alpha=0 \
5 gamma -3 beta =0
$$
Since the third equation is the sum of the first two (this comes from the fact that your inequality is homogeneous) your system has infinitely many solutions. Solving you get
$$9 alpha =15 beta =25 gamma$$
In particular the following is a solution:
$$alpha=25 \
beta =15 \
gamma=9$$
The solution
By AM-GM you have
$$frac{25frac{x^5}{y^3}+15frac{y^5}{z^3}+9frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{25}(frac{y^5}{z^3})^{15}(frac{z^5}{x^3})^{9}}=x^2\
frac{9frac{x^5}{y^3}+25frac{y^5}{z^3}+15frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{9}(frac{y^5}{z^3})^{25}(frac{z^5}{x^3})^{15}}=y^2\
frac{15frac{x^5}{y^3}+9frac{y^5}{z^3}+25frac{z^5}{x^3}}{49}geq sqrt[49]{(frac{x^5}{y^3})^{15}(frac{y^5}{z^3})^{9}(frac{z^5}{x^3})^{25}}=z^2\
$$
Add them together.
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
answered Jan 12 at 3:19
N. S.N. S.
103k6112208
103k6112208
1
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
add a comment |
1
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
1
1
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
Thank you very much for providing the motivation
$endgroup$
– user574848
Jan 12 at 3:20
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
$begingroup$
thank you so much for answering
$endgroup$
– Hoàng
Jan 12 at 3:37
add a comment |
$begingroup$
By AM-GM $$2x^5+3y^5geq5sqrt[5]{(x^5)^2(y^5)^3}=5x^2y^3,$$ which gives
$$frac{x^5}{y^3}geqfrac{5x^2-3y^2}{2}.$$
Id est, $$sum_{cyc}frac{x^5}{y^3}geqsum_{cyc}frac{5x^2-3y^2}{2}=sum_{cyc}x^2.$$
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
2
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
add a comment |
$begingroup$
By AM-GM $$2x^5+3y^5geq5sqrt[5]{(x^5)^2(y^5)^3}=5x^2y^3,$$ which gives
$$frac{x^5}{y^3}geqfrac{5x^2-3y^2}{2}.$$
Id est, $$sum_{cyc}frac{x^5}{y^3}geqsum_{cyc}frac{5x^2-3y^2}{2}=sum_{cyc}x^2.$$
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
2
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
add a comment |
$begingroup$
By AM-GM $$2x^5+3y^5geq5sqrt[5]{(x^5)^2(y^5)^3}=5x^2y^3,$$ which gives
$$frac{x^5}{y^3}geqfrac{5x^2-3y^2}{2}.$$
Id est, $$sum_{cyc}frac{x^5}{y^3}geqsum_{cyc}frac{5x^2-3y^2}{2}=sum_{cyc}x^2.$$
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
By AM-GM $$2x^5+3y^5geq5sqrt[5]{(x^5)^2(y^5)^3}=5x^2y^3,$$ which gives
$$frac{x^5}{y^3}geqfrac{5x^2-3y^2}{2}.$$
Id est, $$sum_{cyc}frac{x^5}{y^3}geqsum_{cyc}frac{5x^2-3y^2}{2}=sum_{cyc}x^2.$$
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 5:30
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
2
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
add a comment |
2
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
2
2
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
Michael.Very nice.in 3 lines!
$endgroup$
– Peter Szilas
Jan 12 at 7:19
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
$begingroup$
@Peter Szila It's just one line: the last line with remark: By AM-GM.
$endgroup$
– Michael Rozenberg
Jan 12 at 7:20
add a comment |
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