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xzlbq

#1 h Dec 26, 2009, 5:13 AM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

Let $x,y,z,u,v,w$ are positive numbers,prove that:

$(x+y+z)(u+v+w)(\frac{1}{(ux+vy+wz}+\frac{1}{uy+vz+wx}+\frac{1}{uz+vx+wy}) \geq9.$
BQ

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kuing

#2 h Dec 26, 2009, 5:17 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

Let $x,y,z,u,v,w$ are positive numbers,prove that:

$(x + y + z)(u + v + w)(\frac {1}{(ux + vy + wz} + \frac {1}{uy + vz + wx} + \frac {1}{uz + vx + wy}) \geq9.$
BQ

yes, that's so easy

$\frac {1}{ux + vy + wz} + \frac {1}{uy + vz + wx} + \frac {1}{uz + vx + wy}$

$\ge \frac{9}{ux + vy + wz+uy + vz + wx+uz + vx + wy}$

$=\frac{9}{(x+y+z)(u+v+w)}$

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xzlbq

#3 h Dec 26, 2009, 5:18 AM • 2 Y

Y by Adventure10, Mango247

(x1+x2+x3+x4)*(a+b+c+d)*(1/(a*x1+b*x2+c*x3+d*x4)+1/(a*x2+b*x3+c*x4+d*x1)+1/(a*x3+b*x4+c*x1+d*x2)+1/(a*x4+b*x1+c*x2+d*x3))-16>=0;;
?
BQ

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kuing

#4 h Dec 26, 2009, 5:26 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

(x1+x2+x3+x4)*(a+b+c+d)*(1/(a*x1+b*x2+c*x3+d*x4)+1/(a*x2+b*x3+c*x4+d*x1)+1/(a*x3+b*x4+c*x1+d*x2)+1/(a*x4+b*x1+c*x2+d*x3))-16>=0;;
?
BQ

the same way.

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xzlbq

#5 h Dec 26, 2009, 6:28 AM • 2 Y

Y by Adventure10, Mango247

replace it:

Let x,y,z,u,v,w are positive numbers,prove that:

$\frac {x}{y + w} + \frac {y}{z + u} + \frac {z}{x + v} + \frac {u}{v + z} + \frac {v}{w + x} + \frac {w}{u + y} \geq 3$

BQ

$\frac{x}{y+w+u}+\frac{y}{z+u+v}+\frac{z}{x+v+w}+\frac{u}{v+z+x}+\frac{v}{w+x+y}+\frac{w}{u+y+z} \geq 2;$

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xzlbq

#6 h Dec 26, 2009, 6:46 AM • 2 Y

Y by Adventure10, Mango247

easy?

$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{u}+\frac{1}{v}+\frac{1}{w}) \geq 9(\frac{1}{ux+vy+wz}+\frac{1}{uy+vz+wx}+\frac{1}{uz+vx+wy})$

BQ

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xzlbq

#7 h Dec 26, 2009, 6:58 AM • 2 Y

Y by Adventure10, Mango247

?
$16\, \left( {\it x_1}\,a + {\it x_2}\,b + {\it x_3}\,c + {\it x_4}\,d \right) ^{ - 1} + 16\, \left( {\it x_2}\,a + {\it x_3}\,b + {\it x_4}\,c + {\it x_1}\,d \right) ^{ - 1} + 16\, \left( {\it x_3}\,a + {\it x_4}\,b + {\it x_1}\,c + {\it x_2 }\,d \right) ^{ - 1} + 16\, \left( {\it x_4}\,a + {\it x_1}\,b + {\it x_2}\,c + { \it x_3}\,d \right) ^{ - 1}\leq \left( {{\it x_1}}^{ - 1} + {{\it x_2}}^{ - 1} + { {\it x_3}}^{ - 1} + {{\it x_4}}^{ - 1} \right) \left( {a}^{ - 1} + {b}^{ - 1} + {c}^{ - 1} + {d}^{ - 1} \right)$

This post has been edited 1 time. Last edited by xzlbq, Dec 26, 2009, 6:59 AM

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kuing

#8 h Dec 26, 2009, 6:58 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

replace it:

Let x,y,z,u,v,w are positive numbers,prove that:

$\frac {x}{y + w} + \frac {y}{z + u} + \frac {z}{x + v} + \frac {u}{v + z} + \frac {v}{w + x} + \frac {w}{u + y} \geq 3$

BQ

$\frac {x}{y + w + u} + \frac {y}{z + u + v} + \frac {z}{x + v + w} + \frac {u}{v + z + x} + \frac {v}{w + x + y} + \frac {w}{u + y + z} \geq 2;$

the frist ineq I try to use the similar way to prove:

$\frac {x}{y + w} + \frac {y}{z + u} + \frac {z}{x + v} + \frac {u}{v + z} + \frac {v}{w + x} + \frac {w}{u + y}$

$= \frac {x^2}{x(y + w)} + \frac {y^2}{y(z + u)} + \frac {z^2}{z(x + v)} + \frac {u^2}{u(v + z)} + \frac {v^2}{v(w + x)} + \frac {w^2}{w(u + y)}$

$\ge \frac {(x + y + z + u + v + w)^2}{x(y + w) + y(z + u) + z(x + v) + u(v + z) + v(w + x) + w(u + y)},$

just need to prove that

$(x + y + z + u + v + w)^2 \ge 3(x(y + w) + y(z + u) + z(x + v) + u(v + z) + v(w + x) + w(u + y)),$

this ineq holds, but I can prove it,

help.......

and the second ineq, still use above way, that's OK:

$\frac {x}{y + w + u} + \frac {y}{z + u + v} + \frac {z}{x + v + w} + \frac {u}{v + z + x} + \frac {v}{w + x + y} + \frac {w}{u + y + z}$

$=\frac {x^2}{x(y + w + u)} + \frac {y^2}{y(z + u + v)} + \frac {z^2}{z(x + v + w)} + \frac {u^2}{u(v + z + x)} + \frac {v^2}{v(w + x + y)} + \frac {w^2}{w(u + y + z)}$

$\ge \frac {(x + y + z + u + v + w)^2}{x(y+w+u)+y(z+u+v)+z(x+v+w)+u(v+z+x)+v(w+x+y)+w(u+y+z)},$

so we just need to prove that

$(x + y + z + u + v + w)^2 \ge 2(x(y+w+u)+y(z+u+v)+z(x+v+w)+u(v+z+x)+v(w+x+y)+w(u+y+z))$

$\iff (x-u)^2+(y-v)^2+(z-w)^2 \ge 0,$

clear true.

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xzlbq

#9 h Dec 26, 2009, 7:14 AM • 2 Y

Y by Adventure10, Mango247

try:
$0\leq {\frac {{\it x_1}}{{\it x_2} + c + d + a}} + {\frac {{\it x_2}}{{\it x_3} + d + a + b}} + {\frac {{\it x_3}}{{\it x_4} + a + b + c}} + {\frac {{\it x_4}}{{\it x_1} + b + c + d}} + {\frac {a}{b + {\it x_3} + {\it x_4} + {\it x_1}}} + {\frac {b}{c + {\it x_4} + {\it x_1} + {\it x_2}}} + {\frac {c}{d + {\it x_1} + {\it x_2} + {\it x_3}}} + {\frac { d}{a + {\it x_2} + {\it x_3} + {\it x_4}}} - 2,$

inside $x_1,x_2,x_3,x_4,a,b,c,d > 0$

back:
tet is:
x1/(x2+c+d+a)+x2/(x3+d+a+b)+x3/(x4+a+b+c)+x4/(x1+b+c+d)+a/(b+x3+x4+x1)+b/(c+x4+x1+x2)+c/(d+x1+x2+x3)+d/(a+x2+x3+x4)-2>=0;;
BQ

This post has been edited 2 times. Last edited by xzlbq, Dec 26, 2009, 7:18 AM

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arqady

#10 h Dec 26, 2009, 7:15 AM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

just need to prove that

$(x + y + z + u + v + w)^2 \ge 3(x(y + w) + y(z + u) + z(x + v) + u(v + z) + v(w + x) + w(u + y)),$

this ineq holds, but I can prove it,

help.......

It's equivalent to $(x+u-w-z)^2+(x+u-y-v)^2+(y+v-w-z)^2\geq0.$

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kuing

#11 h Dec 26, 2009, 7:15 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

easy?

$(\frac {1}{x} + \frac {1}{y} + \frac {1}{z})(\frac {1}{u} + \frac {1}{v} + \frac {1}{w}) \geq 9(\frac {1}{ux + vy + wz} + \frac {1}{uy + vz + wx} + \frac {1}{uz + vx + wy})$

BQ

yes, that's easy

by carlson ineq, we have

$(\frac {1}{x} + \frac {1}{y} + \frac {1}{z})(\frac {1}{u} + \frac {1}{v} + \frac {1}{w})(ux + vy + wz)\ge (1 + 1 + 1)^3,$

so we have

$(\frac {1}{x} + \frac {1}{y} + \frac {1}{z})(\frac {1}{u} + \frac {1}{v} + \frac {1}{w})\ge \frac {27}{ux + vy + wz},$

similarly, we get

$(\frac {1}{x} + \frac {1}{y} + \frac {1}{z})(\frac {1}{u} + \frac {1}{v} + \frac {1}{w})\ge \frac {27}{uy + vz + wx},$

$(\frac {1}{x} + \frac {1}{y} + \frac {1}{z})(\frac {1}{u} + \frac {1}{v} + \frac {1}{w})\ge \frac {27}{uz + vx + wy},$

adding them together, we get the result.

7#'s ineq is similar.

This post has been edited 1 time. Last edited by kuing, Dec 26, 2009, 10:39 AM

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kuing

#12 h Dec 26, 2009, 7:29 AM • 1 Y

Y by Adventure10

arqady wrote:

kuing wrote:

just need to prove that

$(x + y + z + u + v + w)^2 \ge 3(x(y + w) + y(z + u) + z(x + v) + u(v + z) + v(w + x) + w(u + y)),$

this ineq holds, but I can prove it,

help.......

It's equivalent to $(x + u - w - z)^2 + (x + u - y - v)^2 + (y + v - w - z)^2\geq0.$

thank you very much!

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kuing

#13 h Dec 26, 2009, 7:44 AM • 1 Y

Y by Adventure10

xzlbq wrote:

try:
$0\leq {\frac {{\it x_1}}{{\it x_2} + c + d + a}} + {\frac {{\it x_2}}{{\it x_3} + d + a + b}} + {\frac {{\it x_3}}{{\it x_4} + a + b + c}} + {\frac {{\it x_4}}{{\it x_1} + b + c + d}} + {\frac {a}{b + {\it x_3} + {\it x_4} + {\it x_1}}} + {\frac {b}{c + {\it x_4} + {\it x_1} + {\it x_2}}} + {\frac {c}{d + {\it x_1} + {\it x_2} + {\it x_3}}} + {\frac { d}{a + {\it x_2} + {\it x_3} + {\it x_4}}} - 2,$

inside $x_1,x_2,x_3,x_4,a,b,c,d > 0$

back:
tet is:
x1/(x2+c+d+a)+x2/(x3+d+a+b)+x3/(x4+a+b+c)+x4/(x1+b+c+d)+a/(b+x3+x4+x1)+b/(c+x4+x1+x2)+c/(d+x1+x2+x3)+d/(a+x2+x3+x4)-2>=0;;
BQ

similar way, just need to prove that

$( { x_1} + { x_2} + { x_3} + { x_4} + a + b + c + d ) ^{2} \ge 2({ x_1} ( { x_2} + c + d + a ) + { x_2} ( { x_3} + d + a + b ) + { x_3} ( { x_4} + a + b + c ) + { x_4} ( { x_1} + b + c + d ) + a ( b + { x_3} + { x_4} + { x_1} ) + b ( c + { x_4} + { x_1} + { x_2} ) + c ( d + { x_1} + { x_2} + { x_3} ) + d ( a + { x_2} + { x_3} + { x_4} )),$

how about S.O.S this ineq?

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kuing

#14 h Dec 26, 2009, 8:17 AM • 1 Y

Y by Adventure10

kuing wrote:

xzlbq wrote:

try:
$0\leq {\frac {{\it x_1}}{{\it x_2} + c + d + a}} + {\frac {{\it x_2}}{{\it x_3} + d + a + b}} + {\frac {{\it x_3}}{{\it x_4} + a + b + c}} + {\frac {{\it x_4}}{{\it x_1} + b + c + d}} + {\frac {a}{b + {\it x_3} + {\it x_4} + {\it x_1}}} + {\frac {b}{c + {\it x_4} + {\it x_1} + {\it x_2}}} + {\frac {c}{d + {\it x_1} + {\it x_2} + {\it x_3}}} + {\frac { d}{a + {\it x_2} + {\it x_3} + {\it x_4}}} - 2,$

inside $x_1,x_2,x_3,x_4,a,b,c,d > 0$

back:
tet is:
x1/(x2+c+d+a)+x2/(x3+d+a+b)+x3/(x4+a+b+c)+x4/(x1+b+c+d)+a/(b+x3+x4+x1)+b/(c+x4+x1+x2)+c/(d+x1+x2+x3)+d/(a+x2+x3+x4)-2>=0;;
BQ

similar way, just need to prove that

$( { x_1} + { x_2} + { x_3} + { x_4} + a + b + c + d ) ^{2} \ge 2({ x_1} ( { x_2} + c + d + a ) + { x_2} ( { x_3} + d + a + b ) + { x_3} ( { x_4} + a + b + c ) + { x_4} ( { x_1} + b + c + d ) + a ( b + { x_3} + { x_4} + { x_1} ) + b ( c + { x_4} + { x_1} + { x_2} ) + c ( d + { x_1} + { x_2} + { x_3} ) + d ( a + { x_2} + { x_3} + { x_4} )),$

how about S.O.S this ineq?

I got it

$\iff ( a + b + c + d - {x_1} - {x_2} - {x_3} - {x_4}) ^{2} + ( - a + b - c + d + {x_1} - {x_2} + {x_3} - {x_4}) ^{2} \ge 0$

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xzlbq

#15 h Dec 26, 2009, 8:24 AM • 1 Y

Y by Adventure10

If this,must be To the n-variables
BQ

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kuing

#16 h Dec 26, 2009, 10:32 AM • 1 Y

Y by Adventure10

xzlbq wrote:

If this,must be To the n-variables
BQ

I will try to reseach it, but I think, for n-variables, my above way should be hard work, the last S.O.S is so hard to transformation...

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#17 h Jan 1, 2010, 9:42 PM • 1 Y

Y by Adventure10

kuing wrote:

xzlbq wrote:

If this,must be To the n-variables
BQ

I will try to reseach it, but I think, for n-variables, my above way should be hard work, the last S.O.S is so hard to transformation...

The method with C-S doesn't work for $n\ge 5$ . Let $s = \sum x_i$ and $t = \sum y_i$ . For the ineq
$\sum_{cyc}\left(\frac {x_i}{x_{i + 1} + t - y_{i + 1}} + \frac {y_i}{y_{i + 1} + s - x_{i + 1}}\right)\ge2$ which is the general version of the second one in post #5 ( $n = 3$ ) and of post #9 ( $n = 4$ ), after C-S, we would have to prove, putting $z_i: = x_i - y_i\in\mathbb R$ ,
$(\sum z_i)^2\ge 2\sum z_iz_{i + 1}$ But this is false for $n\ge5$ , e.g. $(z_i) = (0,\dots,0,1,1,0, - 2)$ .

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xzlbq

#18 h Sep 12, 2017, 5:44 AM • 2 Y

Y by Adventure10, Mango247

spanferkel wrote:

kuing wrote:

xzlbq wrote:

If this,must be To the n-variables
BQ

I will try to reseach it, but I think, for n-variables, my above way should be hard work,
the last S.O.S is so hard to transformation...

The method with C-S doesn't work for $n\ge 5$ . Let $s = \sum x_i$ and $t = \sum y_i$ . For the ineq
$\sum_{cyc}\left(\frac {x_i}{x_{i + 1} + t - y_{i + 1}} + \frac {y_i}{y_{i + 1} + s - x_{i + 1}}\right)\ge2$ which is the general version of the second one in post #5 ( $n = 3$ ) and of post #9 ( $n = 4$ ),
after C-S, we would have to prove, putting $z_i: = x_i - y_i\in\mathbb R$ ,
$(\sum z_i)^2\ge 2\sum z_iz_{i + 1}$ But this is false for $n\ge5$ , e.g. $(z_i) = (0,\dots,0,1,1,0, - 2)$ .

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