MT【209】打破对称
设正数$a,b,c$满足$ab+bc+ca=47$,求$(a^2+5)(b^2+5)(c^2+5)$的最小值_____
解:$(a^2+5)(b^2+5)(c^2+5)=
(a^2+5)(5(b+c)^2+(bc-5)^2)\ge(\sqrt{5}a(b+c)+\sqrt{5}(bc-5))^2
=5(ab+bc+ca-5)^2=5*42^2=8820$
当$a=5,b=4,c=3$时取到最小值.
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设正数$a,b,c$满足$ab+bc+ca=47$,求$(a^2+5)(b^2+5)(c^2+5)$的最小值_____
解:$(a^2+5)(b^2+5)(c^2+5)=
(a^2+5)(5(b+c)^2+(bc-5)^2)\ge(\sqrt{5}a(b+c)+\sqrt{5}(bc-5))^2
=5(ab+bc+ca-5)^2=5*42^2=8820$
当$a=5,b=4,c=3$时取到最小值.