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Find this inequality by finding the generalizating one too
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)find the min of this inequalityProve this inequality $a^4+b^4+c^4+9ge 4(a^2b+b^2c+c^2a)$How does this inequality imply this one?Proving inequality too simple?Strange inequality over the positive realsProve this inequality $3(abc+4)ge 5(ab+bc+ca)$prove this inequality by $abc=1$How to prove this inequality ???How to prove the given inequality using CS or AM GM HM inequalityto prove this inequality $sum a^3b+3ge 2(ab+bc+ca)$
$begingroup$
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
asked Mar 23 at 12:09
HaiDangelHaiDangel
967
$endgroup$
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
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
asked Mar 23 at 12:09
HaiDangelHaiDangel
967
$endgroup$
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
inequality substitution constants uvw buffalo-way
asked Mar 23 at 12:09
HaiDangelHaiDangel
967
asked Mar 23 at 12:09
HaiDangelHaiDangel
967
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
967
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
$begingroup$
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
1
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
$begingroup$
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 |
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$begingroup$
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