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Question-188387




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

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