Inequality 11

(Crux Mathematicorum)
Cho $a,b,c,x,y,z>0$. CMR:
$\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\ge \frac{3(xy+yz+zx)}{x+y+z}(\ast )$
                                                                Lời Giải
Ta có: $(\ast )\iff \sum [(\frac{a}{b+c}+1)(y+z)]\ge \frac{3(xy+yz+zx)}{x+y+z}+2(x+y+z)$
$\iff \frac{1}{2}[(\Sigma b+c)(\Sigma \frac{y+z}{b+c})]\ge \frac{3(xy+yz+zx)}{x+y+z}+2(x+y+z)$
Áp dụng bất đẳng thức $CBS$:
$\frac{1}{2}(\Sigma b+c)(\Sigma \frac{y+z}{b+c})\ge \frac{1}{2}(\Sigma \sqrt{y+z}{)}^2$
Cần CM: $\frac{1}{2}(\Sigma \sqrt{y+z}{)}^2\ge \frac{3(xy+yz+zx)}{x+y+z}+2(x+y+z)$
$\iff \sum \sqrt{(y+z)(z+x)}\ge \frac{3(xy+yz+zx)}{x+y+z}+(x+y+z)$
Ta có BĐT phụ sau:
$\sum \sqrt{(x+y)(y+z)}\ge \sqrt{(x+y+z{)}^2+9(xy+yz+zx)}(\ast \ast )$
$\iff x^2+y^2+z^2+3(xy+yz+zx)+2[\Sigma (x+y)\sqrt{(y+z)(z+x)}]\ge x^2+y^2+z^2+11(xy+yz+zx)$
$\iff \sum (x+y)\sqrt{(y+z)(z+x)}\ge 4(xy+yz+zx)$
Áp dụng bất đẳng thức $CBS,AM-GM$:
$\sum (x+y)\sqrt{(y+z)(z+x)}\ge \sum (x+y)(\sqrt{xy}+z)=zx+yz+\sqrt{xy}(x+y)$
$\ge \sum xy+yz+2xy=4(xy+yz+zx)\Rightarrow (\ast \ast )$ được CM.
Áp dụng, đặt $x+y+z=m,xy+yz+zx=n(m,n>0)$ $\Rightarrow m^2\ge 3n$
Ta phải CM: $m\sqrt{m^2+9n}\ge 3n+m^2$ $\iff m^4+9m^{2}n\ge m^4+6m^{2}n+9n^2$ $\iff m^2\ge 3n(\text{True})$
Dấu bằng xảy ra $\iff a=b=c,x=y=z$
Vậy BĐT $(\ast )$ được chứng minh $■$