Cho $a,b,c>0$. CMR: $\frac{a+3c}{a+b}+\frac{c+3a}{c+b}+\frac{4b}{c+a}\geq 6$.

• Cho $a,b,c>0$. CMR: $\frac{a+3c}{a+b}+\frac{c+3a}{c+b}+\frac{4b}{c+a}\geq 6$.

$P+2=\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{2a+4b+2c}{c+a}\geq \frac{16(a+c)^2}{(a+c)^2+4b(a+c)+4ac}+\frac{2a+4b+2c}{c+a}\geq \frac{16(a+c)}{2a+4b+2c}+\frac{2a+4b+2c}{c+a}\geq 8 (Cauchy-schwarz)$

Vậy ta có đpcm

• VT  $=\frac{(a+3c)^2}{(a+3c)(a+b)}+\frac{(c+3a)^2}{(c+3a)(c+b)}+\frac{(4b)^2}{4b(c+a)}$$\overset{\text{B.C.S}}{\ge}\frac{[(a+3c)+(c+3a)+(4b)]^2}{(a+3c)(a+b)+(c+3a)(c+b)+4b(c+a)}$$=\frac{(4a+4b+4c)^2}{a^2+c^2+8ab+8bc+6ca}$
  Mà $(4a+4b+4c)^2=16(a^2+b^2+c^2)+32(ab+bc+ca)$$=6(a^2+c^2)+8(a^2+b^2)+8(b^2+c^2)+2(a^2+c^2)+32(ab+bc+ca)$
  $\overset{\text{Côsi}}{\ge}6(a^2+c^2)+16ab+16bc+4ac+32(ab+bc+ca)=6(a^2+c^2+8ab+8bc+6ca)$
  Suy ra $VT\ge 6$. Dấu $=$ xảy ra tại $a=b=c$.