问题补充：
十万火急 柯西不等式习题 d 求证1/(a-b)+1/(b-c）+1/(c-a)≥9/(a-d)0且满足a+b+c=1 求证a3+b3+c3≥(a2+b2+c2)/30,证明a/（b+2c)+b/(c+2a)+c/(a+2b)≥1注.字母后的数字为上标
答案：
只需证明 （a-d) [1/(a-b)+1/(b-c)+1/(c-d)] >= 9 [1/(a-b) +1/(b-c) +1/(c-d)](a-d)
=[1/(a-b) +1/(b-c) +1/(c-d)](a-b+b-c+c-d)
= [1 + (b-c)/(a-b) + (c-d)/(a-b)] + [1 + (a-b)/(b-c) + (c-d)/(b-c)] + [1 + [(a-b)/(c-d) + (b-c)/(a-d)]
= 3 + [(b-c)/(a-b) + (a-b)/(b-c)] + [(c-d)/(a-b) + (a-b)/(c-d)] + [(c-d)/(b-c) + (b-c)/(c-d)]
≥ 3 + 2 + 2 + 2
= 9 所以 ：1/（a-b)+1/(b-c)+1/(c-d)>=9/(a-d)
3.若a,b,c>0,证明a/（b+2c)+b/(c+2a)+c/(a+2b)≥1
利用Cauchy-Schwarz不等式做
[a/(b+2c)+b/(c+2a)+c/(a+2b)]*(3ab+3bc+3ac)
= [a/(b+2c)+b/(c+2a)+c/(a+2b)]*[a(b+2c)+b(c+2a)+c(a+2b)]
≥(a+b+c)^2
a/(b+2c)+b/(c+2a)+c/(a+2b）≥(a+b+c)^2/(3ab+3bc+3ac)
因为 (a+b+c)^2 ≥ 3ab+3bc+3ac 所以
a/(b+2c)+b/(c+2a)+c/(a+2b)≥1,等号当且仅当 a=b=c时成立
