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Question Number 106315 by pticantor last updated on 04/Aug/20

show that ⊛∀ (x_1 ,x_2 ,.....,x_(n ) )∈R^n (Σ_(k=1) ^n x_k )^2 ≤nΣ_(k=1) ^n x_k ^2 ⊛a,b>0 p(x)=x^n +ax+b=0 could not have more than 3 reals solutions

$$\boldsymbol{{show}}\:\boldsymbol{{that}} \\ $$ $$\circledast\forall\:\left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,.....,\boldsymbol{{x}}_{\boldsymbol{{n}}\:} \right)\in\mathbb{R}^{\boldsymbol{{n}}} \\ $$ $$\left(\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{x}}_{\boldsymbol{{k}}} \right)^{\mathrm{2}} \leqslant\boldsymbol{{n}}\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{x}}_{\boldsymbol{{k}}} ^{\mathrm{2}} \\ $$ $$\circledast\boldsymbol{{a}},\boldsymbol{{b}}>\mathrm{0} \\ $$ $$\boldsymbol{{p}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\boldsymbol{{n}}} +\boldsymbol{{ax}}+\boldsymbol{{b}}=\mathrm{0} \\ $$ $$\boldsymbol{{could}}\:\boldsymbol{{not}}\:\boldsymbol{{have}}\:\boldsymbol{{more}}\:\boldsymbol{{than}}\:\mathrm{3} \\ $$ $$\boldsymbol{{reals}}\:\boldsymbol{{solutions}} \\ $$

Answered by Dwaipayan Shikari last updated on 04/Aug/20

Σ_(k=1) ^n x_k ^2 .Σ_(k=1) ^n 1≥(Σ(x_k .1))^2 {Cauchy schwarz inequality} nΣ_(k=1) ^n x_k ^2 ≥(Σ_(k=1) ^n x_k )^2 {Σ^n (1)=n It means n(x_1 ^2 +x_2 ^2 +x_3 ^2 +x_4 ^2 +....)≥(x_1 +x_2 +x_3 +x_4 +.....)^2 (x_1 ^2 +x_2 ^2 +x_3 ^2 +x_4 ^2 +...)≥(((x_1 +x_2 +x_3 +x_4 +......)/(√n)))^2

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{k}} ^{\mathrm{2}} .\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}\geqslant\left(\Sigma\left({x}_{{k}} .\mathrm{1}\right)\right)^{\mathrm{2}} \left\{{Cauchy}\:{schwarz}\:{inequality}\right\} \\ $$ $${n}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{k}} ^{\mathrm{2}} \geqslant\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{k}} \right)^{\mathrm{2}} \:\:\:\:\:\left\{\overset{{n}} {\sum}\left(\mathrm{1}\right)={n}\right. \\ $$ $$ \\ $$ $${It}\:{means} \\ $$ $${n}\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +....\right)\geqslant\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +.....\right)^{\mathrm{2}} \\ $$ $$\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +...\right)\geqslant\left(\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +......}{\sqrt{{n}}}\right)^{\mathrm{2}} \\ $$

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