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Prove that $sqrt1-x^2 + sqrt1-y^2 + sqrt1-z^2geq 4sqrtfrac3-(x^2+y^2+z^2)5+x^2+y^2+z^2$.
Clash Royale CLAN TAG#URR8PPP
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If $x,y,z in[0,1/2]$, with $x+y+z=1$, then prove that:
$$sqrt1-x^2 + sqrt1-y^2 + sqrt1-z^2geq 4sqrtfrac3-(x^2+y^2+z^2)5+x^2+y^2+z^2$$
OK so... I've tried to square the expression and note that $1-x^2=a$, $1-y^2=b$ etc, but finally I got a remark about a new "face" of that ineq:
$$frac2(sqrtab+sqrtbc+sqrtca)a+b+cgeqfrac8+a+b+c8-(a+b+c)$$
so I appreciate any idea or suggestion..
inequality contest-math substitution holder-inequality uvw
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
asked Jul 26 at 20:28
mathematiciangrade8
263
 |Â
show 1 more comment
up vote
1
down vote
favorite
If $x,y,z in[0,1/2]$, with $x+y+z=1$, then prove that:
$$sqrt1-x^2 + sqrt1-y^2 + sqrt1-z^2geq 4sqrtfrac3-(x^2+y^2+z^2)5+x^2+y^2+z^2$$
OK so... I've tried to square the expression and note that $1-x^2=a$, $1-y^2=b$ etc, but finally I got a remark about a new "face" of that ineq:
$$frac2(sqrtab+sqrtbc+sqrtca)a+b+cgeqfrac8+a+b+c8-(a+b+c)$$
so I appreciate any idea or suggestion..
inequality contest-math substitution holder-inequality uvw
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
asked Jul 26 at 20:28
mathematiciangrade8
263
1
@M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice...
– Cave Johnson
Jul 26 at 20:37
1
@CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate.
– M. Winter
Jul 26 at 20:42
2
@M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title.
– Cave Johnson
Jul 26 at 20:48
1
The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D
– uniquesolution
Jul 26 at 21:00
2
@uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it?
– Brian Tung
Jul 26 at 21:26
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $x,y,z in[0,1/2]$, with $x+y+z=1$, then prove that:
$$sqrt1-x^2 + sqrt1-y^2 + sqrt1-z^2geq 4sqrtfrac3-(x^2+y^2+z^2)5+x^2+y^2+z^2$$
OK so... I've tried to square the expression and note that $1-x^2=a$, $1-y^2=b$ etc, but finally I got a remark about a new "face" of that ineq:
$$frac2(sqrtab+sqrtbc+sqrtca)a+b+cgeqfrac8+a+b+c8-(a+b+c)$$
so I appreciate any idea or suggestion..
inequality contest-math substitution holder-inequality uvw
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
asked Jul 26 at 20:28
mathematiciangrade8
263
If $x,y,z in[0,1/2]$, with $x+y+z=1$, then prove that:
$$sqrt1-x^2 + sqrt1-y^2 + sqrt1-z^2geq 4sqrtfrac3-(x^2+y^2+z^2)5+x^2+y^2+z^2$$
OK so... I've tried to square the expression and note that $1-x^2=a$, $1-y^2=b$ etc, but finally I got a remark about a new "face" of that ineq:
$$frac2(sqrtab+sqrtbc+sqrtca)a+b+cgeqfrac8+a+b+c8-(a+b+c)$$
so I appreciate any idea or suggestion..
inequality contest-math substitution holder-inequality uvw
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
asked Jul 26 at 20:28
mathematiciangrade8
263
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
edited Jul 27 at 2:13
Michael Rozenberg
87.7k1578180
87.7k1578180
asked Jul 26 at 20:28
mathematiciangrade8
263
asked Jul 26 at 20:28
mathematiciangrade8
263
asked Jul 26 at 20:28
mathematiciangrade8
263
263
1
@M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice...
– Cave Johnson
Jul 26 at 20:37
1
@CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate.
– M. Winter
Jul 26 at 20:42
2
@M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title.
– Cave Johnson
Jul 26 at 20:48
1
The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D
– uniquesolution
Jul 26 at 21:00
2
@uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it?
– Brian Tung
Jul 26 at 21:26
 |Â
show 1 more comment
1
@M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice...
– Cave Johnson
Jul 26 at 20:37
1
@CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate.
– M. Winter
Jul 26 at 20:42
2
@M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title.
– Cave Johnson
Jul 26 at 20:48
1
The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D
– uniquesolution
Jul 26 at 21:00
2
@uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it?
– Brian Tung
Jul 26 at 21:26
1
1
@M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice...
– Cave Johnson
Jul 26 at 20:37
@M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice...
– Cave Johnson
Jul 26 at 20:37
1
1
@CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate.
– M. Winter
Jul 26 at 20:42
@CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate.
– M. Winter
Jul 26 at 20:42
2
2
@M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title.
– Cave Johnson
Jul 26 at 20:48
@M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title.
– Cave Johnson
Jul 26 at 20:48
1
1
The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D
– uniquesolution
Jul 26 at 21:00
The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D
– uniquesolution
Jul 26 at 21:00
2
2
@uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it?
– Brian Tung
Jul 26 at 21:26
@uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it?
– Brian Tung
Jul 26 at 21:26
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
By Holder
$$left(sum_cycsqrt1-x^2right)^2sum_cycfrac(3-x)^31-x^2geqleft(sum_cyc(3-x)right)^3=8^3=512.$$
Thus, it's enough to prove that
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2>0$ and does not depend on $w^3$.
Also, $$prod_cyc(1-x^2)=-x^2y^2z^2+1+sum_cyc(x^2y^2-x^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$$ and
$$sum_cyc(3-x)^3(1-y^2)(1-z^2)=sum_cyc(27-27x+9x^2-x^3)(1-y^2-z^2+y^2z^2)=$$
$$=sum_cyc(-x^3y^2z^2+9x^2y^2z^2)+C(u,v^2)w^3+D(u,v^2)=26w^6+C(u,v^2)w^3+D(u,v^2),$$
which says that the inequality
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2$$
is equivalent to $f(w^3)geq0,$ where
$$f(w^3)=-left(26+frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2right)w^6+E(u,v^2)w^3+F(u,v^2),$$ which is a concave function.
Here, $A$, $B,$ $C,$ $D,$ $E$ and $F$ they are functions of $u$ and $v^2$ and don't depend on $w^3$.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens in following cases.
- One of our variables is equal to $frac12.$
Let $z=frac12.$
Hence, $y=frac12-x$, where $0leq xleqfrac12.$
We need to prove that $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39geq0,$$ which is true by AM-GM:
$$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39=$$
$$=39-184x(1-2x)-437x^2(1-2x)^2-106x^3(1-2x)^3=$$
$$=39-92(2x(1-2x))-frac4374(2x(1-2x))^2-frac534(2x(1-2x))^3geq$$
$$geq39-92left(frac2x+1-2x2right)^2-frac4374left(frac2x+1-2x2right)^4-frac534left(frac2x+1-2x2right)^6=$$
$$=39-23-frac43764-frac53256>0;$$
- $w^3=0$.
Let $z=0$ and $y=1-x.$
Hence, $1-xleqfrac12,$ which gives $x=frac12$ and we checked it in the previous case.
- Two variables are equal.
Let $y=x$.
Thus, $z=1-2xleqfrac12,$ which gives $frac14leq xleqfrac12.$
This substitution gives:
$$(3x-1)^2(1-x)(x^3+9x^2+7x-1)geq0,$$ which is obvious.
Done!
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By Holder
$$left(sum_cycsqrt1-x^2right)^2sum_cycfrac(3-x)^31-x^2geqleft(sum_cyc(3-x)right)^3=8^3=512.$$
Thus, it's enough to prove that
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2>0$ and does not depend on $w^3$.
Also, $$prod_cyc(1-x^2)=-x^2y^2z^2+1+sum_cyc(x^2y^2-x^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$$ and
$$sum_cyc(3-x)^3(1-y^2)(1-z^2)=sum_cyc(27-27x+9x^2-x^3)(1-y^2-z^2+y^2z^2)=$$
$$=sum_cyc(-x^3y^2z^2+9x^2y^2z^2)+C(u,v^2)w^3+D(u,v^2)=26w^6+C(u,v^2)w^3+D(u,v^2),$$
which says that the inequality
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2$$
is equivalent to $f(w^3)geq0,$ where
$$f(w^3)=-left(26+frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2right)w^6+E(u,v^2)w^3+F(u,v^2),$$ which is a concave function.
Here, $A$, $B,$ $C,$ $D,$ $E$ and $F$ they are functions of $u$ and $v^2$ and don't depend on $w^3$.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens in following cases.
- One of our variables is equal to $frac12.$
Let $z=frac12.$
Hence, $y=frac12-x$, where $0leq xleqfrac12.$
We need to prove that $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39geq0,$$ which is true by AM-GM:
$$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39=$$
$$=39-184x(1-2x)-437x^2(1-2x)^2-106x^3(1-2x)^3=$$
$$=39-92(2x(1-2x))-frac4374(2x(1-2x))^2-frac534(2x(1-2x))^3geq$$
$$geq39-92left(frac2x+1-2x2right)^2-frac4374left(frac2x+1-2x2right)^4-frac534left(frac2x+1-2x2right)^6=$$
$$=39-23-frac43764-frac53256>0;$$
- $w^3=0$.
Let $z=0$ and $y=1-x.$
Hence, $1-xleqfrac12,$ which gives $x=frac12$ and we checked it in the previous case.
- Two variables are equal.
Let $y=x$.
Thus, $z=1-2xleqfrac12,$ which gives $frac14leq xleqfrac12.$
This substitution gives:
$$(3x-1)^2(1-x)(x^3+9x^2+7x-1)geq0,$$ which is obvious.
Done!
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
add a comment |Â
up vote
1
down vote
accepted
By Holder
$$left(sum_cycsqrt1-x^2right)^2sum_cycfrac(3-x)^31-x^2geqleft(sum_cyc(3-x)right)^3=8^3=512.$$
Thus, it's enough to prove that
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2>0$ and does not depend on $w^3$.
Also, $$prod_cyc(1-x^2)=-x^2y^2z^2+1+sum_cyc(x^2y^2-x^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$$ and
$$sum_cyc(3-x)^3(1-y^2)(1-z^2)=sum_cyc(27-27x+9x^2-x^3)(1-y^2-z^2+y^2z^2)=$$
$$=sum_cyc(-x^3y^2z^2+9x^2y^2z^2)+C(u,v^2)w^3+D(u,v^2)=26w^6+C(u,v^2)w^3+D(u,v^2),$$
which says that the inequality
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2$$
is equivalent to $f(w^3)geq0,$ where
$$f(w^3)=-left(26+frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2right)w^6+E(u,v^2)w^3+F(u,v^2),$$ which is a concave function.
Here, $A$, $B,$ $C,$ $D,$ $E$ and $F$ they are functions of $u$ and $v^2$ and don't depend on $w^3$.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens in following cases.
- One of our variables is equal to $frac12.$
Let $z=frac12.$
Hence, $y=frac12-x$, where $0leq xleqfrac12.$
We need to prove that $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39geq0,$$ which is true by AM-GM:
$$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39=$$
$$=39-184x(1-2x)-437x^2(1-2x)^2-106x^3(1-2x)^3=$$
$$=39-92(2x(1-2x))-frac4374(2x(1-2x))^2-frac534(2x(1-2x))^3geq$$
$$geq39-92left(frac2x+1-2x2right)^2-frac4374left(frac2x+1-2x2right)^4-frac534left(frac2x+1-2x2right)^6=$$
$$=39-23-frac43764-frac53256>0;$$
- $w^3=0$.
Let $z=0$ and $y=1-x.$
Hence, $1-xleqfrac12,$ which gives $x=frac12$ and we checked it in the previous case.
- Two variables are equal.
Let $y=x$.
Thus, $z=1-2xleqfrac12,$ which gives $frac14leq xleqfrac12.$
This substitution gives:
$$(3x-1)^2(1-x)(x^3+9x^2+7x-1)geq0,$$ which is obvious.
Done!
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By Holder
$$left(sum_cycsqrt1-x^2right)^2sum_cycfrac(3-x)^31-x^2geqleft(sum_cyc(3-x)right)^3=8^3=512.$$
Thus, it's enough to prove that
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2>0$ and does not depend on $w^3$.
Also, $$prod_cyc(1-x^2)=-x^2y^2z^2+1+sum_cyc(x^2y^2-x^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$$ and
$$sum_cyc(3-x)^3(1-y^2)(1-z^2)=sum_cyc(27-27x+9x^2-x^3)(1-y^2-z^2+y^2z^2)=$$
$$=sum_cyc(-x^3y^2z^2+9x^2y^2z^2)+C(u,v^2)w^3+D(u,v^2)=26w^6+C(u,v^2)w^3+D(u,v^2),$$
which says that the inequality
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2$$
is equivalent to $f(w^3)geq0,$ where
$$f(w^3)=-left(26+frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2right)w^6+E(u,v^2)w^3+F(u,v^2),$$ which is a concave function.
Here, $A$, $B,$ $C,$ $D,$ $E$ and $F$ they are functions of $u$ and $v^2$ and don't depend on $w^3$.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens in following cases.
- One of our variables is equal to $frac12.$
Let $z=frac12.$
Hence, $y=frac12-x$, where $0leq xleqfrac12.$
We need to prove that $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39geq0,$$ which is true by AM-GM:
$$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39=$$
$$=39-184x(1-2x)-437x^2(1-2x)^2-106x^3(1-2x)^3=$$
$$=39-92(2x(1-2x))-frac4374(2x(1-2x))^2-frac534(2x(1-2x))^3geq$$
$$geq39-92left(frac2x+1-2x2right)^2-frac4374left(frac2x+1-2x2right)^4-frac534left(frac2x+1-2x2right)^6=$$
$$=39-23-frac43764-frac53256>0;$$
- $w^3=0$.
Let $z=0$ and $y=1-x.$
Hence, $1-xleqfrac12,$ which gives $x=frac12$ and we checked it in the previous case.
- Two variables are equal.
Let $y=x$.
Thus, $z=1-2xleqfrac12,$ which gives $frac14leq xleqfrac12.$
This substitution gives:
$$(3x-1)^2(1-x)(x^3+9x^2+7x-1)geq0,$$ which is obvious.
Done!
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
By Holder
$$left(sum_cycsqrt1-x^2right)^2sum_cycfrac(3-x)^31-x^2geqleft(sum_cyc(3-x)right)^3=8^3=512.$$
Thus, it's enough to prove that
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2>0$ and does not depend on $w^3$.
Also, $$prod_cyc(1-x^2)=-x^2y^2z^2+1+sum_cyc(x^2y^2-x^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$$ and
$$sum_cyc(3-x)^3(1-y^2)(1-z^2)=sum_cyc(27-27x+9x^2-x^3)(1-y^2-z^2+y^2z^2)=$$
$$=sum_cyc(-x^3y^2z^2+9x^2y^2z^2)+C(u,v^2)w^3+D(u,v^2)=26w^6+C(u,v^2)w^3+D(u,v^2),$$
which says that the inequality
$$frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2geqsum_cycfrac(3-x)^31-x^2$$
is equivalent to $f(w^3)geq0,$ where
$$f(w^3)=-left(26+frac32(5+x^2+y^2+z^2)3-x^2-y^2-z^2right)w^6+E(u,v^2)w^3+F(u,v^2),$$ which is a concave function.
Here, $A$, $B,$ $C,$ $D,$ $E$ and $F$ they are functions of $u$ and $v^2$ and don't depend on $w^3$.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens in following cases.
- One of our variables is equal to $frac12.$
Let $z=frac12.$
Hence, $y=frac12-x$, where $0leq xleqfrac12.$
We need to prove that $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39geq0,$$ which is true by AM-GM:
$$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39=$$
$$=39-184x(1-2x)-437x^2(1-2x)^2-106x^3(1-2x)^3=$$
$$=39-92(2x(1-2x))-frac4374(2x(1-2x))^2-frac534(2x(1-2x))^3geq$$
$$geq39-92left(frac2x+1-2x2right)^2-frac4374left(frac2x+1-2x2right)^4-frac534left(frac2x+1-2x2right)^6=$$
$$=39-23-frac43764-frac53256>0;$$
- $w^3=0$.
Let $z=0$ and $y=1-x.$
Hence, $1-xleqfrac12,$ which gives $x=frac12$ and we checked it in the previous case.
- Two variables are equal.
Let $y=x$.
Thus, $z=1-2xleqfrac12,$ which gives $frac14leq xleqfrac12.$
This substitution gives:
$$(3x-1)^2(1-x)(x^3+9x^2+7x-1)geq0,$$ which is obvious.
Done!
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
edited Jul 27 at 2:17
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
answered Jul 27 at 2:10
Michael Rozenberg
87.7k1578180
87.7k1578180
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
add a comment |Â
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
Why someone down voted?
– Michael Rozenberg
Jul 27 at 6:49
add a comment |Â
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1
@M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice...
– Cave Johnson
Jul 26 at 20:37
1
@CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate.
– M. Winter
Jul 26 at 20:42
2
@M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title.
– Cave Johnson
Jul 26 at 20:48
1
The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D
– uniquesolution
Jul 26 at 21:00
2
@uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it?
– Brian Tung
Jul 26 at 21:26