How To Prove That For Every Real Number N The Equation A^2+B^2+C^2+D^2=Abc+Abd+Acd+Bcd Has A Solution In Which A,B, C,D Are All Integers Greater Than N

Given $a,b,c,d>0$ & $a^2+b^2+c^2+d^2=1$, prove $$a+b+c+dge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$

The inequality can be written in the condensed form$$sumlimits_Symagesumlimits_Syma^3+sumlimits_Symab$$

I was told that this is a pretty inequality lớn prove, but I have been unable to bởi vì so.

I"ve tried naive things lượt thích multiplying both sides by $a+b+c+d$, và rewriting $(a^2+b^2+c^2+d^2)^2$, but nothing panned out (and the computations were relatively time-consuming). I also tried looking for clever applications of Cauchy-Schwarz (which seems lượt thích the way to lớn go) and AM-GM, but nothing sprung out at me.

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inequality polynomials substitution uvw
giới thiệu
edited Oct 25, 2018 at 7:21

Michael Rozenberg
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asked May 14, năm ngoái at 23:53

Peter WoolfittPeter Woolfitt
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3 Answers 3

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Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $abc+abd+acd+bcd=4w^3$.

Hence, $16u^2-12v^2=1$ và our inequality is equivalent to $3v^6-4uv^2w^3+w^6geq0$.

By Roll"s theorem there are $x>0$, $y>0$ and $z>0$, for which

$x+y+z=3u$, $xy+xz+yz=3v^2$ và $xyz=w^3$.

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After this substitution we need lớn prove that

$sumlimits_cyc(x^3y^3-x^3y^2z-x^3z^2y+x^2y^2z^2)geq0$, which is Schur.

cốt truyện
answered May 17, 2015 at 20:27

Michael RozenbergMichael Rozenberg
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I got some hints from Crux Problem 3059, click here for more details=)

Based on the hints (which tries lớn relate the inequality with a constrained optimization problem), I worked out a proof as follows:

We haveeginalignfrac12(a+b+c+d)^2=½(a^2+b^2+c^2+d^2)+ab+ac+ad+bc+bd+cd\=½+ab+ac+ad+bc+bd+cdendalignby assumption. Thus, in order lớn prove the inequality, it is equivalent khổng lồ prove$$a+b+c+dgeq a^3+b^3+c^3+d^3+frac12(a+b+c+d)^2-frac12.$$This can further be simplified as proving$$a^3+b^3+c^3+d^3+frac12(a+b+c+d-1)^2leq1.~~~~(*)$$Now we try to lớn maximize the LHS under the constraint, i.e.eginalignmax&~~f riangleq a^3+b^3+c^3+d^3+frac12(a+b+c+d-1)^2\s.t.&~~a^2+b^2+c^2+d^2=1.endalignNow we try lớn use the Lagrangian multiplier method. Let$$L=f+lambda(a^2+b^2+c^2+d^2-1).$$Take the derivative of $L$ regarding $a,b,c,d$ respectively và let the derivative equal khổng lồ $0$ gives the following set of equations:eginalignL_a=&a+b+c+d+3a^2+2alambda=1\L_b=&a+b+c+d+3b^2+2blambda=1\L_c=&a+b+c+d+3c^2+2clambda=1\L_d=&a+b+c+d+3d^2+2dlambda=1,endalign

Notice that these four equations giới thiệu exactly the same form. Thus, either$$3x^2+2lambda x=0,~~x=a,b,c,d,~~~~(case1)$$or$$a=b=c=d.~~~~(case2)$$We analyse these two cases separately:

Case 1:Under this case, we have$$a+b+c+d=1.$$Since$$a^2+b^2+c^2+d^2=1,$$the only possibility is one of four elements $a,b,c,d$ equal to $1$ and others are all $0$, which gives $lambda=-3/2$ and $f=1$.

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Case 2:By the assumption$$a^2+b^2+c^2+d^2=1,$$it follows$$a=b=c=d=frac12,$$which gives $lambda=-7/4$ and $f=1$.