Question Number 188387 by cortano12 last updated on 28/Feb/23
Answered by Frix last updated on 28/Feb/23
$$\mathrm{I}\:\mathrm{think}\:\mathrm{because}\:\mathrm{of}\:\mathrm{symmetry}\:{a}={b}={c}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{Minimum}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by mnjuly1970 last updated on 28/Feb/23
$$ \\ $$$$\:\:{E}=\:\frac{\frac{\mathrm{1}}{{a}^{\:\mathrm{2}} }}{{ab}+{ac}}\:+\frac{\frac{\mathrm{1}}{{b}^{\:\mathrm{2}} }}{{ab}\:+{bc}}\:+\frac{\frac{\mathrm{1}}{{c}^{\:\mathrm{2}} }}{{ac}\:+\:{bc}} \\ $$$$\:\:\:\overset{{T}_{\mathrm{2}−} {lemma}} {\geqslant}\:\:\frac{\left(\:\frac{\mathrm{1}}{{a}}\:+\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\right)^{\:\mathrm{2}} }{\mathrm{2}\left({ab}\:+{ac}\:+{bc}\right)} \\ $$$$\:\:\:\:\:\:\:\overset{{abc}=\mathrm{1}} {=}\:\frac{\left(\:{ab}\:+\:{ac}\:+{bc}\right)^{\:\mathrm{2}} }{\mathrm{2}\left({ab}\:+\:{ac}\:+{bc}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{ab}\:+{ac}\:+{bc}\:\overset{{am}−{gm}} {\geqslant}\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{{a}^{\:\mathrm{2}} {b}^{\:\mathrm{2}} \:{c}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:{min}\:\left(\:{E}\:\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$