求证:1⼀2a+1⼀2b+1⼀2c大于1⼀(b+c)+1⼀(a+c)+1⼀(a+b)

1/4a+1/4b
=(a+b)/4ab
≥(a+b)/(a+b)^2
=1/(a+b)
同理1/4b+1/4c≥1/(b+c)
1/4c+1/4a≥1/(c+a)
由以上三式可得1/2a+1/2b+1/2c≥1/(a+b)+1/(b+c)+1/(c+a)

1/a+1/b≥2/√ab≥2/[(b+a)/2]=4/(b+a)(此处两个不等号均用了不等式x+y≥2√xy)
从而1/4a+1/4b≥1/(b+a)
同理1/4a+1/4c≥1/(c+a)
1/4b+1/4c≥1/(c+b)
相加得到1/2a+1/2b+1/2c≥1/(b+c)+1/(a+c)+1/(a+b)
由于a,b,c互不相等,故等号取不到
1/2a+1/2b+1/2c＞1/(b+c)+1/(a+c)+1/(a+b)

1/a+1/b≥2/√ab≥2/[(b+a)/2]=4/(b+a)(此处两个不等号均用了不等式x+y≥2√xy)
从而1/4a+1/4b≥1/(b+a)
同理1/4a+1/4c≥1/(c+a)
1/4b+1/4c≥1/(c+b)
相加得到1/2a+1/2b+1/2c≥1/(b+c)+1/(a+c)+1/(a+b)
由于a,b,c互不相等,故等号取不到
1/2a+1/2b+1/2c＞1/(b+c)+1/(a+c)+1/(a+b)