Áp dụng bất đẳng thức Cauchy dạng `1/x+1/y≥4/(x+y)` cho `a;b;c` là các số thực dương ta có:
$*$`1/(a+3b)+1/(b+2c+a)≥(4)/((a+3b)+(b+2c+a))=4/(2a+4b+2c)=4/(2(a+2b+c))=2/(a+2b+c)`
$*$`1/(b+3c)+1/(c+2a+b)≥4/((b+3c)+(c+2a+b))=4/(2a+2b+4c)=4/(2(a+b+2c))=2/(b+2c+a)`
$*$`1/(c+3a)+1/(a+2b+c)≥4/((c+3a)+(a+2b+c))=4/(4a+2b+2c)=4/(2(2a+b+c))=2/(c+2a+b)`
`⇒(1/(a+3b)+1/(b+2c+a))+(1/(b+3c)+1/(c+2a+b))+(1/(c+3a)+1/(a+2b+c))≥2/(a+2b+c)+2/(b+2c+a)+2/(c+2a+b)`
`⇔1/(a+3b)+1/(b+3c)+1/(c+3a)≥2/(a+2b+c)+2/(b+2c+a)+2/(c+2a+b)-1/(a+2b+c)-1/(b+2c+a)-1/(c+2a+b)`
`⇔1/(a+3b)+1/(b+3c)+1/(c+3a)≥1/(a+2b+c)+1/(b+2c+a)+1/(c+2a+b)`
`⇔1/(a+2b+c)+1/(b+2c+a)+1/(c+2a+b)≤1/(a+3b)+1/(b+3c)+1/(c+3a)`
Dấu "=" xảy ra khi và chỉ khi:
$\begin{cases}a+3b=b+2c+a\\b+3c=c+2a+b\\c+3a=a+2b+c\end{cases}$
$⇔a=b=c$
