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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edited Oct 25, 2018 at 7:21

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

Peter WoolfittPeter Woolfitt
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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.

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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 haveeginalignfrac12(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\$.