Cho a,b,c là các số dương. Chứng minh rằng: `\frac{4}{a+2b+c}` + `\frac{4}{2a+b+c}` + `\frac{4}{a+b+2c}` `>=` `\frac{9}{a+b+c}`

Ta nhận thấy:

`\frac{4}{a+2b+c}=\frac{2^2}{a+2b+c}`

Tương tự:

`\frac{4}{2a+b+c}=\frac{2^2}{2x+b+c}`

`\frac{4}{a+b+2c}=\frac{2^2}{a+b+2c}`

Áp dụng bất đẳng thức  Schwarz, ta có:

`\frac{4}{a+2b+c}+\frac{4}{2a+b+c}+\frac{4}{a+b+2c}\ge \frac{(2+2+2)^2}{a+2b+c+2a+b+c+a+b+2c}`

`<=> \frac{4}{a+2b+c}+\frac{4}{2a+b+c}+\frac{4}{a+b+2c} \ge \frac{36}{4a+4b+4c}`

`<=> \frac{4}{a+2b+c}+\frac{4}{2a+b+c}+\frac{4}{a+b+2c} \ge \frac{36}{4(a+b+c)}`

`<=> \frac{4}{a+2b+c}+\frac{4}{2a+b+c}+\frac{4}{a+b+2c} \ge \frac{9}{a+b+c} \ đpcm`

Dâu `=` xảy ra khi: `\frac{2}{a+2b+c}=\frac{2}{2a+b+c}=\frac{4}{a+b+2c}`

`=> a+2b+c=2a+b+c=a+b+2c`

`=> a=b=c`