Bạn trên làm cách THCS nha bạn, chắc do dùng kí hiệu nhiều quá nên bạn không hiểu.
Ta có: [tex]\frac{1}{a+b+1}+\frac{1}{3}\geq \frac{4}{a+b+4}[/tex](BĐT phụ [tex]\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}[/tex])
Tương tự thì [tex]\left\{\begin{matrix} \frac{1}{b+c+1}+\frac{1}{3}\geq \frac{4}{b+c+4}\\ \frac{1}{c+a+1}+\frac{1}{3}\geq \frac{4}{c+a+4} \end{matrix}\right.\Rightarrow 4(\frac{1}{a+b+4}+\frac{1}{b+c+4}+\frac{1}{c+a+4})\leq \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}+3.\frac{1}{3}=\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}+1\Rightarrow \frac{1}{a+b+4}+\frac{1}{b+c+4}+\frac{1}{c+a+4}\leq \frac{1}{4}(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}+1)[/tex]
Bây giờ đặt [tex]a=x^3,b=y^3,c=z^3\Rightarrow xyz=\sqrt[3]{abc}=1[/tex]
Ta có: [tex]\frac{1}{a+b+1}=\frac{1}{x^3+y^3+xyz}[/tex]
Vì [tex]x^3+y^3=(x+y)(x^2+y^2-xy)\geq (x+y)(2xy-xy)=xy(x+y)\Rightarrow \frac{1}{x^3+y^3+xyz}\leq \frac{1}{xy(x+y)+xyz}=\frac{1}{xy(x+y+z)}=\frac{z}{xyz(x+y+z)}=\frac{z}{x+y+z}[/tex]
Tương tự thì [tex]\frac{1}{b+c+1}=\frac{1}{y^3+z^3+xyz}\geq \frac{x}{x+y+z},\frac{1}{c+a+1}=\frac{1}{z^3+x^3+xyz}\geq \frac{y}{x+y+z}\Rightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}+1\geq \frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}+1=2\Rightarrow \frac{1}{a+b+4}+\frac{1}{b+c+4}+\frac{1}{c+a+4}\leq \frac{1}{4}(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}+1)\leq \frac{1}{4}.2=\frac{1}{2}(đpcm)[/tex]