Question-221195

Question Number 221195 by Rojarani last updated on 26/May/25

Commented by Ghisom last updated on 26/May/25

$$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{49}}{\mathrm{34}} \\ $$

Answered by mr W last updated on 27/May/25

(a/(b+c))+1+(b/(c+a))+1+(c/(a+b))+1=35 (a+b+c)((1/(a+b))+(1/(b+c))+(1/(c+a)))=35 (a+b+c)×5=35 ⇒a+b+c=7 (b+c)(c+a)+(a+b)(c+a)+(a+b)(b+c)=5(a+b)(b+c)(c+a) (7−a)(7−b)+(7−b)(7−c)+(7−c)(7−a)=5(7−a)(7−b)(7−c) 3×7^2 −2×7(a+b+c)+ab+bc+ca=5[7^3 −(a+b+c)×7^2 +7(ab+bc+ca)−abc] 3×7^2 −2×7^2 +ab+bc+ca=5[7^3 −7×7^2 +7(ab+bc+ca)−abc] ⇒ab+bc+ca=((49+5abc)/(34))≥((49)/(34)) i.e. (ab+bc+ca)_(min) =((49)/(34))

$$\frac{{a}}{{b}+{c}}+\mathrm{1}+\frac{{b}}{{c}+{a}}+\mathrm{1}+\frac{{c}}{{a}+{b}}+\mathrm{1}=\mathrm{35} \\ $$$$\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}+{b}}+\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}\right)=\mathrm{35} \\ $$$$\left({a}+{b}+{c}\right)×\mathrm{5}=\mathrm{35} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{7} \\ $$$$\left({b}+{c}\right)\left({c}+{a}\right)+\left({a}+{b}\right)\left({c}+{a}\right)+\left({a}+{b}\right)\left({b}+{c}\right)=\mathrm{5}\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right) \\ $$$$\left(\mathrm{7}−{a}\right)\left(\mathrm{7}−{b}\right)+\left(\mathrm{7}−{b}\right)\left(\mathrm{7}−{c}\right)+\left(\mathrm{7}−{c}\right)\left(\mathrm{7}−{a}\right)=\mathrm{5}\left(\mathrm{7}−{a}\right)\left(\mathrm{7}−{b}\right)\left(\mathrm{7}−{c}\right) \\ $$$$\mathrm{3}×\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{7}\left({a}+{b}+{c}\right)+{ab}+{bc}+{ca}=\mathrm{5}\left[\mathrm{7}^{\mathrm{3}} −\left({a}+{b}+{c}\right)×\mathrm{7}^{\mathrm{2}} +\mathrm{7}\left({ab}+{bc}+{ca}\right)−{abc}\right] \\ $$$$\mathrm{3}×\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{7}^{\mathrm{2}} +{ab}+{bc}+{ca}=\mathrm{5}\left[\mathrm{7}^{\mathrm{3}} −\mathrm{7}×\mathrm{7}^{\mathrm{2}} +\mathrm{7}\left({ab}+{bc}+{ca}\right)−{abc}\right] \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\frac{\mathrm{49}+\mathrm{5}{abc}}{\mathrm{34}}\geqslant\frac{\mathrm{49}}{\mathrm{34}} \\ $$$${i}.{e}.\:\left({ab}+{bc}+{ca}\right)_{{min}} =\frac{\mathrm{49}}{\mathrm{34}} \\ $$

Commented by Ghisom last updated on 27/May/25

c=7−(a+b) (1/(a+b))+(1/(b+c))+(1/(c+a))=5 ⇒ b=((7−a)/2)±((√((a+7)(5a^2 −3a−210)))/(2(√(5a−34)))) c=((7−a)/2)∓((√((a+7)(5a^2 −3a−210)))/(2(√(5a−34)))) ⇒ ab+bc+ca=−((5a^3 −35a^2 +49)/(5a−34))=f(a) a, b, c ≥0: max (f(a)) =((61(√(4209))−3667)/(200)); a=((67−(√(4209)))/(20)) min (f(a)) =((49)/(34)); a=0∨a=(7/2)±((7(√(255)))/(34))

$${c}=\mathrm{7}−\left({a}+{b}\right) \\ $$$$\frac{\mathrm{1}}{{a}+{b}}+\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}=\mathrm{5} \\ $$$$\Rightarrow \\ $$$${b}=\frac{\mathrm{7}−{a}}{\mathrm{2}}\pm\frac{\sqrt{\left({a}+\mathrm{7}\right)\left(\mathrm{5}{a}^{\mathrm{2}} −\mathrm{3}{a}−\mathrm{210}\right)}}{\mathrm{2}\sqrt{\mathrm{5}{a}−\mathrm{34}}} \\ $$$${c}=\frac{\mathrm{7}−{a}}{\mathrm{2}}\mp\frac{\sqrt{\left({a}+\mathrm{7}\right)\left(\mathrm{5}{a}^{\mathrm{2}} −\mathrm{3}{a}−\mathrm{210}\right)}}{\mathrm{2}\sqrt{\mathrm{5}{a}−\mathrm{34}}} \\ $$$$\Rightarrow \\ $$$${ab}+{bc}+{ca}=−\frac{\mathrm{5}{a}^{\mathrm{3}} −\mathrm{35}{a}^{\mathrm{2}} +\mathrm{49}}{\mathrm{5}{a}−\mathrm{34}}={f}\left({a}\right) \\ $$$${a},\:{b},\:{c}\:\geqslant\mathrm{0}: \\ $$$$\mathrm{max}\:\left({f}\left({a}\right)\right)\:=\frac{\mathrm{61}\sqrt{\mathrm{4209}}−\mathrm{3667}}{\mathrm{200}};\:{a}=\frac{\mathrm{67}−\sqrt{\mathrm{4209}}}{\mathrm{20}} \\ $$$$\mathrm{min}\:\left({f}\left({a}\right)\right)\:=\frac{\mathrm{49}}{\mathrm{34}};\:{a}=\mathrm{0}\vee{a}=\frac{\mathrm{7}}{\mathrm{2}}\pm\frac{\mathrm{7}\sqrt{\mathrm{255}}}{\mathrm{34}} \\ $$

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