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`
