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信息發(fā)布者:
$a + 3b - c$
4
12.3
2
14
9
解:
(1)當(dāng)$n = 4$時(shí),有$(2,3,3);$
當(dāng)$n = 5$時(shí),有$(2,4,4),$$(3,3,4);$
當(dāng)$n = 6$時(shí),有$(2,5,5),$$(3,4,5),$$(4,4,4).$
(2)當(dāng)$n = 12$時(shí),$a + b + c = 24,$且$a + b>c,$$a\leqslant b\leqslant c,$由此得$8\leqslant c\leqslant11,$即$c = 8,9,10,11.$
故可得$(a,b,c)$共有12組,分別為$(2,11,11),$$(3,10,11),$$(4,9,11),$$(5,8,11),$$(6,7,11),$$(4,10,10),$$(5,9,10),$$(6,8,10),$$(7,7,10),$$(6,9,9),$$(7,8,9),$$(8,8,8).$

證明:
(1)延長(zhǎng)$BD$交$AC$于$E,$在$\triangle ABE$中,有$AB + AE>BE,$在$\triangle EDC$中,有$ED + EC>CD,$$\therefore AB + AE+ED + EC>BE + CD.$$\because AE + EC = AC,$$BE = BD + DE,$$\therefore AB + AC+ED>BD + DE+CD,$$\therefore BD + CD<AB + AC.$
(2)由
(1)同理可得:$AB + BC>AD + CD,$$BC + AC>BD + AD,$$AB + AC>BD + CD,$$\therefore 2(AB + BC + AC)>2(AD + BD + CD),$$\therefore AB + BC + AC>AD + BD + CD.$