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Question Number 171681 by Shrinava last updated on 19/Jun/22

Answered by mindispower last updated on 19/Jun/22

We Have Ln increase function witch is octave function ⇔aln(cot(A))+bln(cot(B))+cln(cot(C))≤(a+b+c)ln(((3R)/(a+b+c))) (1/(a+b+c))(aln(cot(A))+bln(Cot(B))+cln(cot(C))≤ ln(((acot(A))/(a+b+c))+((bcot(B))/(a+b+c))+((cCot(C))/(a+b+c)))....E =ln((1/(a+b+c))(((acos(A))/(sin(A)))+((bCos(B))/(sin(B)))+((cCos(C))/(sin(C)))) Sinus Relation (a/(sin(A)))=(b/(sin(B)))=(c/(sin(C)))=2R LH≤ln(((2R)/(a+b+c))(cos(A)+cos(B)+cos(C)) cos is concave ⇒cos(A)+cos(B)+cos(C)≤3cos(((A+B+C)/3)) =(3/2) ⇔LH≤ln(((2R)/(a+b+c)).(3/2))=ln(((3R)/(a+b+c))) (1/(a+b+c))(aln(cot(A))+bln(Cot(B))+cln(cot(C))≤ln(((3R)/(a+b+c))) ⇒aln(cot(A)+bln(cot(B))+cln(cot(C))≤ln((((3R)/(a+b+c)))^(a+b+c) ) ⇔cot^a (A)cot^b (B)cot^c (C)≤(((3R)/(a+b+c)))^((a+b+c))

$${We}\:{Have}\:{Ln}\:{increase}\:{function}\:\:{witch}\:{is}\:{octave}\:{function} \\ $$$$\Leftrightarrow{aln}\left({cot}\left({A}\right)\right)+{bln}\left({cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant\left({a}+{b}+{c}\right){ln}\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right) \\ $$$$\frac{\mathrm{1}}{{a}+{b}+{c}}\left({aln}\left({cot}\left({A}\right)\right)+{bln}\left({Cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant\right. \\ $$$${ln}\left(\frac{{acot}\left({A}\right)}{{a}+{b}+{c}}+\frac{{bcot}\left({B}\right)}{{a}+{b}+{c}}+\frac{{cCot}\left({C}\right)}{{a}+{b}+{c}}\right)....{E} \\ $$$$={ln}\left(\frac{\mathrm{1}}{{a}+{b}+{c}}\left(\frac{{acos}\left({A}\right)}{{sin}\left({A}\right)}+\frac{{bCos}\left({B}\right)}{{sin}\left({B}\right)}+\frac{{cCos}\left({C}\right)}{{sin}\left({C}\right)}\right)\right. \\ $$$${Sinus}\:{Relation}\:\frac{{a}}{{sin}\left({A}\right)}=\frac{{b}}{{sin}\left({B}\right)}=\frac{{c}}{{sin}\left({C}\right)}=\mathrm{2}{R} \\ $$$${LH}\leqslant{ln}\left(\frac{\mathrm{2}{R}}{{a}+{b}+{c}}\left({cos}\left({A}\right)+{cos}\left({B}\right)+{cos}\left({C}\right)\right)\right. \\ $$$${cos}\:{is}\:{concave}\:\Rightarrow{cos}\left({A}\right)+{cos}\left({B}\right)+{cos}\left({C}\right)\leqslant\mathrm{3}{cos}\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Leftrightarrow{LH}\leqslant{ln}\left(\frac{\mathrm{2}{R}}{{a}+{b}+{c}}.\frac{\mathrm{3}}{\mathrm{2}}\right)={ln}\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right) \\ $$$$\frac{\mathrm{1}}{{a}+{b}+{c}}\left({aln}\left({cot}\left({A}\right)\right)+{bln}\left({Cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant{ln}\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right)\right. \\ $$$$\Rightarrow{aln}\left({cot}\left({A}\right)+{bln}\left({cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant{ln}\left(\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right)^{{a}+{b}+{c}} \right)\right. \\ $$$$\Leftrightarrow{cot}^{{a}} \left({A}\right){cot}^{{b}} \left({B}\right){cot}^{{c}} \left({C}\right)\leqslant\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right)^{\left({a}+{b}+{c}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Shrinava last updated on 20/Jun/22

Cool dear professor thank you

$$\mathrm{Cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mindispower last updated on 20/Jun/22

withe pleasur

$${withe}\:{pleasur} \\ $$

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