Question Number 155272 by bekzodjumayev last updated on 28/Sep/21
Commented by bekzodjumayev last updated on 28/Sep/21
$$\boldsymbol{{P}}{rove}\:{that} \\ $$$${Please}\:{help} \\ $$
Commented by puissant last updated on 28/Sep/21
![alna+blnb+clnc ≥(a+b+c)ln(((a+b+c)/3)) ≥((a+b+c)/3) [lna+lnb+lnc] ≥ ((a+b+c)/3) [ln(abc)] ⇒ ln(a^a b^b c^c )≥ln[(abc)^((a+b+c)/3) ] ⇒ a^a b^b c^c ≥ (abc)^((a+b+c)/3) ...](https://www.tinkutara.com/question/Q155289.png)
$$\:\:\:\:{alna}+{blnb}+{clnc} \\ $$$$\geqslant\left({a}+{b}+{c}\right){ln}\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right) \\ $$$$\geqslant\frac{{a}+{b}+{c}}{\mathrm{3}}\:\left[{lna}+{lnb}+{lnc}\right] \\ $$$$\geqslant\:\frac{{a}+{b}+{c}}{\mathrm{3}}\:\left[{ln}\left({abc}\right)\right] \\ $$$$\Rightarrow\:{ln}\left({a}^{{a}} {b}^{{b}} {c}^{{c}} \right)\geqslant{ln}\left[\left({abc}\right)^{\frac{{a}+{b}+{c}}{\mathrm{3}}} \right] \\ $$$$ \\ $$$$\Rightarrow\:{a}^{{a}} {b}^{{b}} {c}^{{c}} \:\geqslant\:\left({abc}\right)^{\frac{{a}+{b}+{c}}{\mathrm{3}}} … \\ $$
Commented by bekzodjumayev last updated on 28/Sep/21
$${Thank}\:{you} \\ $$