Cho 1/a + 1/b + 1/c = 16 Chứng minh : 1/3a + 2b+c + 1/a+3b+2c + 1/2a+3c+b ≤ 8/3

Đáp án:

Áp dụng bđt cosi-schwart cho 6 số ta có:

`1/a+1/a+1/a+1/b+1/b+1/c>=36/(3a+2b+c)`

`<=>1/(3a+2b+c)<=1/36(1/a+1/a+1/a+1/b+1/b+1/c)=1/36(3/a+2/b+1/c)`

Chứng minh tương tự ta có:

`1/(a+3b+2c)<=1/36(3/b+2/c+1/a)`

`1/(2a+3c+b)<=1/36(3/c+2/a+1/b)`

`=>VT<=1/36(6/a+6/b+6/c)=1/6(1/a+1/b+1/c)=8/3.`

Dấu “=” xảy ra khi `a=b=c=3/16`