Question-211207

Question Number 211207 by Etimbuk last updated on 31/Aug/24

Answered by mr W last updated on 03/Sep/24

e_1 =a+b+c+d e_2 =ab+bc+cd+da+ac+bd e_3 =abc+bcd+cda+dab e_4 =abcd S_n =a^n +b^n +c^n +d^n S_1 =e_1 =8 S_2 =e_1 S_1 −2e_2 =e_1 ^2 −2e_2 =64−2e_2 S_3 =e_1 S_2 −e_2 S_1 +3e_3 =e_1 ^3 −3e_1 e_2 +3e_3 =512−24e_2 +3e_3 S_4 =e_1 S_3 −e_2 S_2 +e_3 S_1 −4e_4 =e_1 ^4 +2e_2 ^2 −4e_1 ^2 e_2 +4e_1 e_3 −4e_4 =4096+2e_2 ^2 −256e_2 +32e_3 −4e_4 Φ=20S_2 −8S_3 +S_4 =20(64−2e_2 )−8(512−24e_2 +3e_3 )+4096+2e_2 ^2 −256e_2 +32e_3 −4e_4 =20×64−40e_2 −8×512+192e_2 −24e_3 +4096+2e_2 ^2 −256e_2 +32e_3 −4e_4 =−72+2(e_2 −26)^2 +4(2e_3 −e_4 ) Φ can take any real value! there is no minimum! there is no maximum!

$${e}_{\mathrm{1}} ={a}+{b}+{c}+{d} \\ $$$${e}_{\mathrm{2}} ={ab}+{bc}+{cd}+{da}+{ac}+{bd} \\ $$$${e}_{\mathrm{3}} ={abc}+{bcd}+{cda}+{dab} \\ $$$${e}_{\mathrm{4}} ={abcd} \\ $$$${S}_{{n}} ={a}^{{n}} +{b}^{{n}} +{c}^{{n}} +{d}^{{n}} \\ $$$${S}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{8} \\ $$$${S}_{\mathrm{2}} ={e}_{\mathrm{1}} {S}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} =\mathrm{64}−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${S}_{\mathrm{3}} ={e}_{\mathrm{1}} {S}_{\mathrm{2}} −{e}_{\mathrm{2}} {S}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{512}−\mathrm{24}{e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$${S}_{\mathrm{4}} ={e}_{\mathrm{1}} {S}_{\mathrm{3}} −{e}_{\mathrm{2}} {S}_{\mathrm{2}} +{e}_{\mathrm{3}} {S}_{\mathrm{1}} −\mathrm{4}{e}_{\mathrm{4}} ={e}_{\mathrm{1}} ^{\mathrm{4}} +\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{1}} ^{\mathrm{2}} {e}_{\mathrm{2}} +\mathrm{4}{e}_{\mathrm{1}} {e}_{\mathrm{3}} −\mathrm{4}{e}_{\mathrm{4}} =\mathrm{4096}+\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{256}{e}_{\mathrm{2}} +\mathrm{32}{e}_{\mathrm{3}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$\Phi=\mathrm{20}{S}_{\mathrm{2}} −\mathrm{8}{S}_{\mathrm{3}} +{S}_{\mathrm{4}} \\ $$$$\:\:\:=\mathrm{20}\left(\mathrm{64}−\mathrm{2}{e}_{\mathrm{2}} \right)−\mathrm{8}\left(\mathrm{512}−\mathrm{24}{e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \right)+\mathrm{4096}+\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{256}{e}_{\mathrm{2}} +\mathrm{32}{e}_{\mathrm{3}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$\:\:\:=\mathrm{20}×\mathrm{64}−\mathrm{40}{e}_{\mathrm{2}} −\mathrm{8}×\mathrm{512}+\mathrm{192}{e}_{\mathrm{2}} −\mathrm{24}{e}_{\mathrm{3}} +\mathrm{4096}+\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{256}{e}_{\mathrm{2}} +\mathrm{32}{e}_{\mathrm{3}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$\:\:\:=−\mathrm{72}+\mathrm{2}\left({e}_{\mathrm{2}} −\mathrm{26}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{2}{e}_{\mathrm{3}} −{e}_{\mathrm{4}} \right) \\ $$$$\Phi\:{can}\:{take}\:{any}\:{real}\:{value}! \\ $$$${there}\:{is}\:{no}\:{minimum}! \\ $$$${there}\:{is}\:{no}\:{maximum}! \\ $$

Answered by AlagaIbile last updated on 03/Sep/24

$$\mathrm{112} \\ $$

Commented by mr W last updated on 03/Sep/24

$${please}\:{show}\:{how}\:{you}\:{got}\:{this}. \\ $$

Commented by AlagaIbile last updated on 08/Sep/24

Σ_(cyclic) a^4 − 8a^3 + 20a^2 = Σ_(cyclic) [(a^2 − 4a + 2)^2 + 16a − 4] = Σ_(cyclic) (a^2 − 4a + 2)^2 + 16(8) − 4(4) ≥ 112 Equality is attained when a = b = 2 + (√2) and c = d = 2 − (√2)

$$\:\:\underset{{cyclic}} {\sum}\:\boldsymbol{{a}}^{\mathrm{4}} \:−\:\mathrm{8}\boldsymbol{{a}}^{\mathrm{3}} \:+\:\mathrm{20}\boldsymbol{{a}}^{\mathrm{2}} \\ $$$$\:=\:\underset{{cyclic}} {\sum}\left[\left(\boldsymbol{{a}}^{\mathrm{2}} \:−\:\mathrm{4}\boldsymbol{{a}}\:+\:\mathrm{2}\right)^{\mathrm{2}} +\:\mathrm{16}\boldsymbol{{a}}\:−\:\mathrm{4}\right] \\ $$$$\:=\:\underset{\boldsymbol{{cyclic}}} {\sum}\left(\boldsymbol{{a}}^{\mathrm{2}} \:−\:\mathrm{4}\boldsymbol{{a}}\:+\:\mathrm{2}\right)^{\mathrm{2}} +\:\mathrm{16}\left(\mathrm{8}\right)\:−\:\mathrm{4}\left(\mathrm{4}\right)\:\geqslant\:\mathrm{112} \\ $$$$\:{Equality}\:{is}\:{attained}\:{when}\:\boldsymbol{{a}}\:=\:\boldsymbol{{b}}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{2}} \\ $$$$\:{and}\:\boldsymbol{{c}}\:=\:\boldsymbol{{d}}\:=\:\mathrm{2}\:−\:\sqrt{\mathrm{2}} \\ $$

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