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Question-151294

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Question Number 151294 by Jonathanwaweh last updated on 19/Aug/21
Answered by ArielVyny last updated on 20/Aug/21
(a+b+c)=1→a^2 +a(b+c)+b(a+c)+b^2 +c^2 +c(b+a) a^2 +b^2 +c^2 +2ab+2ac+2bc=1 A=(1/2)ac.sin(b^� )=(1/2)bc.sin(a^� )=(1/2)ab.sin(c^� ) c^2 =a^2 +b^2 −2a.b.cos(c^� ) a^2 =c^2 +b^2 −2c.b.cos(a^� ) →alkashi′s theorem b^2 =a^2 +c^2 −2a.c.cos(b^� ) a^2 +b^2 +c^2 =2a^2 +2b^2 +2c^2 −2(a.b.cos(c^� )+c.b.cos(a^� )+a.c.cos(b^� )) a^2 +b^2 +c^2 >0 2a^2 +2b^2 +2c^2 −2(a.b.cos(c^� )+c.b.cos(a^� )+a.c.cos(b^� ))=1−2(ab+ac+bc)>0 par identification 2a^2 +2b^2 +2c^2 <1→a^2 +b^2 +c^2 <(1/2) (∗)
$$\left({a}+{b}+{c}\right)=\mathrm{1}\rightarrow{a}^{\mathrm{2}} +{a}\left({b}+{c}\right)+{b}\left({a}+{c}\right)+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{c}\left({b}+{a}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{bc}=\mathrm{1} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}{ac}.{sin}\left(\hat {{b}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{bc}.{sin}\left(\hat {{a}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ab}.{sin}\left(\hat {{c}}\right) \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}.{b}.{cos}\left(\hat {{c}}\right) \\ $$$${a}^{\mathrm{2}} ={c}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}.{b}.{cos}\left(\hat {{a}}\right)\:\:\rightarrow{alkashi}'{s}\:{theorem} \\ $$$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{a}.{c}.{cos}\left(\hat {{b}}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −\mathrm{2}\left({a}.{b}.{cos}\left(\hat {{c}}\right)+{c}.{b}.{cos}\left(\hat {{a}}\right)+{a}.{c}.{cos}\left(\hat {{b}}\right)\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −\mathrm{2}\left({a}.{b}.{cos}\left(\hat {{c}}\right)+{c}.{b}.{cos}\left(\hat {{a}}\right)+{a}.{c}.{cos}\left(\hat {{b}}\right)\right)=\mathrm{1}−\mathrm{2}\left({ab}+{ac}+{bc}\right)>\mathrm{0} \\ $$$${par}\:{identification}\: \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} <\mathrm{1}\rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} <\frac{\mathrm{1}}{\mathrm{2}}\:\left(\ast\right) \\ $$

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