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Question Number 177068 by LOSER last updated on 30/Sep/22
Let a,b,c be real numbers such that: a+b+c=0. Prove that: ((a^2 b^2 c^2 )/4)+(((ab+bc+ca)^3 )/(27))≤0
Leta,b,cberealnumberssuchthat:a+b+c=0.Provethat:a2b2c24+(ab+bc+ca)327⩽0
Answered by Frix last updated on 30/Sep/22
c=−a−b ((a^2 b^2 c^2 )/4)+(((ab+bc+ca)^3 )/(27))= =−(((a−b)^2 (a+2b)^2 (2a+b)^2 )/(108))≤0 obviously true
c=−a−ba2b2c24+(ab+bc+ca)327==−(a−b)2(a+2b)2(2a+b)2108⩽0obviouslytrue
Commented by Frix last updated on 30/Sep/22
how I got this: c=−a−b∧b=pa ⇒ c=−a(1+p) ((a^2 b^2 c^2 )/4)+(((ab+bc+ca)^3 )/(27))= =−((a^6 (p^6 +3p^5 −((3p^4 )/4)−((13p^3 )/2)−((3p^2 )/4)+3p+1))/(27)) factors of the polynome p^6 +3p^5 −((3p^4 )/4)−((13p^3 )/2)−((3p^2 )/4)+3p+1 p=q−(1/2) q^6 −((9q^4 )/2)+((81q^2 )/(16)) q^2 (q^2 −(9/4))^2 q^2 (q−(3/2))^2 (q+(3/2))^2 q=p+(1/2) (p−1)^2 (p+2)^2 (p+(1/2))^2 p=(b/a) (((a−b)^2 (a+2b)^2 (2a+b)^2 )/(4a^6 ))
howIgotthis:c=−a−b∧b=pa⇒c=−a(1+p)a2b2c24+(ab+bc+ca)327==−a6(p6+3p5−3p44−13p32−3p24+3p+1)27factorsofthepolynomep6+3p5−3p44−13p32−3p24+3p+1p=q−12q6−9q42+81q216q2(q2−94)2q2(q−32)2(q+32)2q=p+12(p−1)2(p+2)2(p+12)2p=ba(a−b)2(a+2b)2(2a+b)24a6
Commented by LOSER last updated on 01/Oct/22
You′re too strong, sir!
You′retoostrong,sir!

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