Question Number 189269 by Rupesh123 last updated on 14/Mar/23
Answered by a.lgnaoui last updated on 15/Mar/23
![posons:E=(1/a)+(1/b)+(1/c)+((15)/(a+b+c)) (a+b+c)×E=15+(a+b+c)((1/a)+(1/b)+(1/c)) =18+((a(b+c))/(bc))+((b(a+c))/(ac))+((c(a+b))/(ab)) =18+(((1−2abc−ab)c+ab(a+b))/(abc)) =18+(((1−2abc−ab)/(ab)) +((a+b)/c)) (i) 2abc+(a+b)c+ab=1 (a>0,b>0,c>0) 1−2abc−ab=(a+b)c { ((E(a+b+c)=18+(((a+b)c)/(ab))+((a+b)/c))),((E=((18)/((a+b+c)))+(([(a+b+c)−c)]c)/(ab(a+b+c)))+(((a+b+c)−c)/(c(a+b+c))))) :} posons ( a+b+c)=λ ((18)/λ)+(((λ−c)c)/(λab))+((λ−c)/(λc))=((18)/λ)+((λ−c)/λ)((c/(ab))+(1/c)) =(1/λ)[18+((λ/c)−1)(1+(c^2 /(ab)))] =(1/λ)[17+((((a+b)c)/(ab))+(λ/c) ) =((17abc)/((a+b+c)))+(((a+b)c)/((a+b+c)ab))+(1/c) pour tout a b c>0 abc>(a+b+c) ⇒((abc)/λ)>1 (((a+b)c)/(λab))+(1/c)>0⇒E≥17 ⇒(1/a)+(1/b)+(1/c)+((15)/(a+b+c))≥16 i](https://www.tinkutara.com/question/Q189385.png)
$${posons}:{E}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{15}}{{a}+{b}+{c}} \\ $$$$\left({a}+{b}+{c}\right)×{E}=\mathrm{15}+\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right) \\ $$$$=\mathrm{18}+\frac{{a}\left({b}+{c}\right)}{{bc}}+\frac{{b}\left({a}+{c}\right)}{{ac}}+\frac{{c}\left({a}+{b}\right)}{{ab}} \\ $$$$=\mathrm{18}+\frac{\left(\mathrm{1}−\mathrm{2}{abc}−{ab}\right){c}+{ab}\left({a}+{b}\right)}{{abc}} \\ $$$$=\mathrm{18}+\left(\frac{\mathrm{1}−\mathrm{2}{abc}−{ab}}{{ab}}\:\:+\frac{{a}+{b}}{{c}}\right)\:\:\left({i}\right) \\ $$$$\mathrm{2}{abc}+\left({a}+{b}\right){c}+{ab}=\mathrm{1}\:\:\:\left({a}>\mathrm{0},{b}>\mathrm{0},{c}>\mathrm{0}\right) \\ $$$$\mathrm{1}−\mathrm{2}{abc}−{ab}=\left({a}+{b}\right){c} \\ $$$$\begin{cases}{{E}\left({a}+{b}+{c}\right)=\mathrm{18}+\frac{\left({a}+{b}\right){c}}{{ab}}+\frac{{a}+{b}}{{c}}}\\{{E}=\frac{\mathrm{18}}{\left({a}+{b}+{c}\right)}+\frac{\left.\left[\left({a}+{b}+{c}\right)−{c}\right)\right]{c}}{{ab}\left({a}+{b}+{c}\right)}+\frac{\left({a}+{b}+{c}\right)−{c}}{{c}\left({a}+{b}+{c}\right)}}\end{cases} \\ $$$${posons}\:\left(\:{a}+{b}+{c}\right)=\lambda \\ $$$$\frac{\mathrm{18}}{\lambda}+\frac{\left(\lambda−{c}\right){c}}{\lambda{ab}}+\frac{\lambda−{c}}{\lambda{c}}=\frac{\mathrm{18}}{\lambda}+\frac{\lambda−{c}}{\lambda}\left(\frac{{c}}{{ab}}+\frac{\mathrm{1}}{{c}}\right) \\ $$$$=\frac{\mathrm{1}}{\lambda}\left[\mathrm{18}+\left(\frac{\lambda}{{c}}−\mathrm{1}\right)\left(\mathrm{1}+\frac{{c}^{\mathrm{2}} }{{ab}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\lambda}\left[\mathrm{17}+\left(\frac{\left({a}+{b}\right){c}}{{ab}}+\frac{\lambda}{{c}}\:\right)\right. \\ $$$$=\frac{\mathrm{17}{abc}}{\left({a}+{b}+{c}\right)}+\frac{\left({a}+{b}\right){c}}{\left({a}+{b}+{c}\right){ab}}+\frac{\mathrm{1}}{{c}} \\ $$$${pour}\:{tout}\:\:{a}\:{b}\:{c}>\mathrm{0}\:\:\:\:{abc}>\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow\frac{{abc}}{\lambda}>\mathrm{1}\:\:\:\frac{\left({a}+{b}\right){c}}{\lambda{ab}}+\frac{\mathrm{1}}{{c}}>\mathrm{0}\Rightarrow\boldsymbol{{E}}\geqslant\mathrm{17} \\ $$$$ \\ $$$$\Rightarrow\frac{\mathrm{1}}{\boldsymbol{{a}}}+\frac{\mathrm{1}}{\boldsymbol{{b}}}+\frac{\mathrm{1}}{\boldsymbol{{c}}}+\frac{\mathrm{15}}{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}\geqslant\mathrm{16} \\ $$$${i} \\ $$