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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
$${Let}\:{a},{b},{c}\:{be}\:{real}\:{numbers}\:{such}\:{that}: \\ $$$${a}+{b}+{c}=\mathrm{0}.\:{Prove}\:{that}: \\ $$$$\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{4}}+\frac{\left({ab}+{bc}+{ca}\right)^{\mathrm{3}} }{\mathrm{27}}\leqslant\mathrm{0} \\ $$
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
$${c}=−{a}−{b} \\ $$$$\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{4}}+\frac{\left({ab}+{bc}+{ca}\right)^{\mathrm{3}} }{\mathrm{27}}= \\ $$$$=−\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+\mathrm{2}{b}\right)^{\mathrm{2}} \left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} }{\mathrm{108}}\leqslant\mathrm{0}\:\mathrm{obviously}\:\mathrm{true} \\ $$
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 ))
$$\mathrm{how}\:\mathrm{I}\:\mathrm{got}\:\mathrm{this}: \\ $$$${c}=−{a}−{b}\wedge{b}={pa}\:\Rightarrow\:{c}=−{a}\left(\mathrm{1}+{p}\right) \\ $$$$\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{4}}+\frac{\left({ab}+{bc}+{ca}\right)^{\mathrm{3}} }{\mathrm{27}}= \\ $$$$=−\frac{{a}^{\mathrm{6}} \left({p}^{\mathrm{6}} +\mathrm{3}{p}^{\mathrm{5}} −\frac{\mathrm{3}{p}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{13}{p}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{3}{p}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{p}+\mathrm{1}\right)}{\mathrm{27}} \\ $$$$\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynome} \\ $$$${p}^{\mathrm{6}} +\mathrm{3}{p}^{\mathrm{5}} −\frac{\mathrm{3}{p}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{13}{p}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{3}{p}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{p}+\mathrm{1} \\ $$$${p}={q}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${q}^{\mathrm{6}} −\frac{\mathrm{9}{q}^{\mathrm{4}} }{\mathrm{2}}+\frac{\mathrm{81}{q}^{\mathrm{2}} }{\mathrm{16}} \\ $$$${q}^{\mathrm{2}} \left({q}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$${q}^{\mathrm{2}} \left({q}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \left({q}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${q}={p}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({p}−\mathrm{1}\right)^{\mathrm{2}} \left({p}+\mathrm{2}\right)^{\mathrm{2}} \left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${p}=\frac{{b}}{{a}} \\ $$$$\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+\mathrm{2}{b}\right)^{\mathrm{2}} \left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{6}} } \\ $$
Commented by LOSER last updated on 01/Oct/22
You′re too strong, sir!
$${You}'{re}\:{too}\:{strong},\:{sir}! \\ $$

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