∵a+b>c, a²-ab+b²=(a-b/2)²+3b²/4>0∴(a+b-c)(a²-ab+b²)=(a+b)(a²-ab+b²)-c(a²+b²-ab)=a³+b³-c(a²+b²+2ab-3ab)=a³+b³+3abc-(a+b)²c>0∴a³+b³+3abc>(a+b)²c∴a³+b³+c³+3abc>c³+(a+b)²c=c[c²+(a+b)²]>c×2√c²(a+b)²=2(a+b)c²
