Gọi O là giao điểm AC và BD
Do lăng trụ đều \(\Rightarrow AC\perp\left(BDD'B'\right)\Rightarrow AC\perp\left(EOF\right)\)
\(V_{ACEF}=V_{AOEF}+V_{COEF}=2V_{AOEF}=\dfrac{2}{3}AO.S_{OEF}=\dfrac{a\sqrt{2}}{3}.S_{OEF}\)
Đặt \(BE=x;\) \(DF=y\), trên BB' lấy G sao cho \(BG=DF=y\)
\(\Rightarrow FG=BD=a\sqrt{2}\) và \(EG=\left|x-y\right|\)
\(\Rightarrow EF=\sqrt{EG^2+FG^2}=\sqrt{2a^2+\left(x-y\right)^2}\)
\(OE=\sqrt{OB^2+BE^2}=\sqrt{\dfrac{a^2}{2}+x^2}\) ; \(OF=\sqrt{OD^2+DF^2}=\sqrt{\dfrac{a^2}{2}+y^2}\)
Do \(\left(EAC\right)\perp\left(FAC\right)\Rightarrow OE\perp OF\)
\(\Rightarrow OE^2+OF^2=EF^2\)
\(\Rightarrow a^2+x^2+y^2=2a^2+\left(x-y\right)^2\Rightarrow xy=\dfrac{a^2}{2}\)
\(S_{OEF}=\dfrac{1}{2}OE.OF=\dfrac{1}{2}\sqrt{\left(\dfrac{a^2}{2}+x^2\right)\left(\dfrac{a^2}{2}+y^2\right)}=\dfrac{1}{2}\sqrt{\dfrac{a^4}{4}+\left(xy\right)^2+\dfrac{a^2}{2}\left(x^2+y^2\right)}\)
\(=\dfrac{1}{2}\sqrt{\dfrac{a^4}{2}+\dfrac{a^2}{2}\left(x^2+y^2\right)}\ge\dfrac{1}{2}\sqrt{\dfrac{a^4}{2}+\dfrac{a^2}{2}.2xy}=\dfrac{1}{2}\sqrt{\dfrac{a^4}{2}+a^2.\dfrac{a^2}{2}}=\dfrac{a^2}{2}\)
\(\Rightarrow V_{ACEF}\ge\dfrac{a\sqrt{2}}{3}.\dfrac{a^2}{2}=\dfrac{a^3\sqrt{2}}{6}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{a\sqrt{2}}{2}\)
