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Find this inequality by finding the generalizating one too

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$begingroup$

Given $x,,y,,z$ such that $x,,y,,zin [,1,,8,]$$.$ Prove$:$
$$frac{x}{y}+ frac{y}{z}+ frac{z}{x}geqq frac{2,x}{y+ z}+ frac{2,y}{z+ x}+ frac{2,z}{x+ y}$$
By computer$,$ I have found the generalization of the above inequality $($If you are interested in$)$$:$
Given $x,,y,,z$ such that $x,,y,,zin [,1,,text{k},]$$.$
$$frac{x}{y}+ frac{y}{z}+ frac{z}{x}geqq frac{2,x}{y+ z}+ frac{2,y}{z+ x}+ frac{2,z}{x+ y}$$
is true for $text{k}> 8$ and its largest value is the largest root of irreducible equation$:$
$$4,text{k}^{,17}- 52,text{k}^{,16}+ 192,text{k}^{,15}- 187,,text{k}^{,14}- 506,text{k}^{,13}- 567,,text{k}^{,12}- 3460,,text{k}^{,11}- 1552,text{k}^{,10}+ 4468,text{k}^{,9}+ 1430,text{k}^{,8}- 2812,text{k}^{,7}- 562,text{k}^{,6}+ 728,text{k}^{,5}- 28,text{k}^{,4}- 156,text{k}^{,3}- 19,text{k}^{,2}+ 6,text{k}+ 1= 0$$
My hard work is substituting $a= frac{x}{y},,b= frac{x}{z}$ $($Suppose $x= max{,x,,y,,z,}$$)$$.$ We will have$:$
$$f(,a,,b,)= a+ frac{b}{a}+ frac{1}{b}- frac{2,ab}{a+ b}- frac{2,b}{ab+ a}- frac{2,a}{ab+ b}geqq 0$$
then$:$
$$f(,a,,8,)= a+ frac{8}{a}+ frac{1}{b}- frac{16,a}{a+ 8}- frac{16}{9,a}- frac{2,a}{8,a+ 8}$$
We have$:$
$$f(,a,,8,)geqq f(,1,,8,)$$
So I tried$:$ $f(,a,,b,)- f(,a,,8,)= (,8- a,),g(,a,,b,)$$,$ then I continued $g(,a,,b,)- g(,a,,8,)$$,$$...$$,$ but $n(,1,,8,)leqq 0$$.$ So I can$'$t continue doing it$!$ Can you try to solve my own$?$ Thanks for your interest$!$

inequality substitution constants uvw buffalo-way

share|cite|improve this question

edited Mar 25 at 23:17

HaiDangel

asked Mar 23 at 12:09

HaiDangelHaiDangel

967

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  These inequalities have no suprema, which would reduce your equation. You are proving the tangent lie group SU(2), which are reducing will heirarch these rationals under an infinum integer resolved by k.
  $endgroup$
  – Cppg
  Mar 23 at 20:44
  

  
  
  
  

add a comment | 

1

2

$begingroup$

Given $x,,y,,z$ such that $x,,y,,zin [,1,,8,]$$.$ Prove$:$
$$frac{x}{y}+ frac{y}{z}+ frac{z}{x}geqq frac{2,x}{y+ z}+ frac{2,y}{z+ x}+ frac{2,z}{x+ y}$$
By computer$,$ I have found the generalization of the above inequality $($If you are interested in$)$$:$
Given $x,,y,,z$ such that $x,,y,,zin [,1,,text{k},]$$.$
$$frac{x}{y}+ frac{y}{z}+ frac{z}{x}geqq frac{2,x}{y+ z}+ frac{2,y}{z+ x}+ frac{2,z}{x+ y}$$
is true for $text{k}> 8$ and its largest value is the largest root of irreducible equation$:$
$$4,text{k}^{,17}- 52,text{k}^{,16}+ 192,text{k}^{,15}- 187,,text{k}^{,14}- 506,text{k}^{,13}- 567,,text{k}^{,12}- 3460,,text{k}^{,11}- 1552,text{k}^{,10}+ 4468,text{k}^{,9}+ 1430,text{k}^{,8}- 2812,text{k}^{,7}- 562,text{k}^{,6}+ 728,text{k}^{,5}- 28,text{k}^{,4}- 156,text{k}^{,3}- 19,text{k}^{,2}+ 6,text{k}+ 1= 0$$
My hard work is substituting $a= frac{x}{y},,b= frac{x}{z}$ $($Suppose $x= max{,x,,y,,z,}$$)$$.$ We will have$:$
$$f(,a,,b,)= a+ frac{b}{a}+ frac{1}{b}- frac{2,ab}{a+ b}- frac{2,b}{ab+ a}- frac{2,a}{ab+ b}geqq 0$$
then$:$
$$f(,a,,8,)= a+ frac{8}{a}+ frac{1}{b}- frac{16,a}{a+ 8}- frac{16}{9,a}- frac{2,a}{8,a+ 8}$$
We have$:$
$$f(,a,,8,)geqq f(,1,,8,)$$
So I tried$:$ $f(,a,,b,)- f(,a,,8,)= (,8- a,),g(,a,,b,)$$,$ then I continued $g(,a,,b,)- g(,a,,8,)$$,$$...$$,$ but $n(,1,,8,)leqq 0$$.$ So I can$'$t continue doing it$!$ Can you try to solve my own$?$ Thanks for your interest$!$

inequality substitution constants uvw buffalo-way

share|cite|improve this question

edited Mar 25 at 23:17

HaiDangel

asked Mar 23 at 12:09

HaiDangelHaiDangel

967

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  These inequalities have no suprema, which would reduce your equation. You are proving the tangent lie group SU(2), which are reducing will heirarch these rationals under an infinum integer resolved by k.
  $endgroup$
  – Cppg
  Mar 23 at 20:44
  

  
  
  
  

add a comment | 

$begingroup$

Given $x,,y,,z$ such that $x,,y,,zin [,1,,8,]$$.$ Prove$:$
$$frac{x}{y}+ frac{y}{z}+ frac{z}{x}geqq frac{2,x}{y+ z}+ frac{2,y}{z+ x}+ frac{2,z}{x+ y}$$
By computer$,$ I have found the generalization of the above inequality $($If you are interested in$)$$:$
Given $x,,y,,z$ such that $x,,y,,zin [,1,,text{k},]$$.$
$$frac{x}{y}+ frac{y}{z}+ frac{z}{x}geqq frac{2,x}{y+ z}+ frac{2,y}{z+ x}+ frac{2,z}{x+ y}$$
is true for $text{k}> 8$ and its largest value is the largest root of irreducible equation$:$
$$4,text{k}^{,17}- 52,text{k}^{,16}+ 192,text{k}^{,15}- 187,,text{k}^{,14}- 506,text{k}^{,13}- 567,,text{k}^{,12}- 3460,,text{k}^{,11}- 1552,text{k}^{,10}+ 4468,text{k}^{,9}+ 1430,text{k}^{,8}- 2812,text{k}^{,7}- 562,text{k}^{,6}+ 728,text{k}^{,5}- 28,text{k}^{,4}- 156,text{k}^{,3}- 19,text{k}^{,2}+ 6,text{k}+ 1= 0$$
My hard work is substituting $a= frac{x}{y},,b= frac{x}{z}$ $($Suppose $x= max{,x,,y,,z,}$$)$$.$ We will have$:$
$$f(,a,,b,)= a+ frac{b}{a}+ frac{1}{b}- frac{2,ab}{a+ b}- frac{2,b}{ab+ a}- frac{2,a}{ab+ b}geqq 0$$
then$:$
$$f(,a,,8,)= a+ frac{8}{a}+ frac{1}{b}- frac{16,a}{a+ 8}- frac{16}{9,a}- frac{2,a}{8,a+ 8}$$
We have$:$
$$f(,a,,8,)geqq f(,1,,8,)$$
So I tried$:$ $f(,a,,b,)- f(,a,,8,)= (,8- a,),g(,a,,b,)$$,$ then I continued $g(,a,,b,)- g(,a,,8,)$$,$$...$$,$ but $n(,1,,8,)leqq 0$$.$ So I can$'$t continue doing it$!$ Can you try to solve my own$?$ Thanks for your interest$!$

inequality substitution constants uvw buffalo-way

share|cite|improve this question

edited Mar 25 at 23:17

HaiDangel

asked Mar 23 at 12:09

HaiDangelHaiDangel

967

$endgroup$

share|cite|improve this question

edited Mar 25 at 23:17

HaiDangel

asked Mar 23 at 12:09

HaiDangelHaiDangel

967

share|cite|improve this question

edited Mar 25 at 23:17

HaiDangel

asked Mar 23 at 12:09

HaiDangelHaiDangel

967

asked Mar 23 at 12:09

HaiDangelHaiDangel

967

asked Mar 23 at 12:09

HaiDangelHaiDangel

967