[tex]\frac{2}{x+\frac{1}{2}\sqrt{x.4y}+\frac{1}{4}\sqrt[3]{x.4y.16z}}\geq \frac{2}{x+\frac{1}{4}(x+y)+\frac{1}{12}(x+y+z)}=\frac{3}{2(x+y+z)}[/tex]
Sau đó đặt [tex]\frac{1}{\sqrt{x+y+z}}=t>0[/tex]
[tex]\frac{2}{x+\frac{1}{2}\sqrt{x.4y}+\frac{1}{4}\sqrt[3]{x.4y.16z}}\geq \frac{2}{x+\frac{1}{4}(x+y)+\frac{1}{12}(x+y+z)}=\frac{3}{2(x+y+z)}[/tex]
Sau đó đặt [tex]\frac{1}{\sqrt{x+y+z}}=t>0[/tex]
Đúng rồi bạn, [tex]\sqrt{ab}\leq \frac{1}{2}(a+b);\, \sqrt[3]{abc}\leq \frac{1}{3}(a+b+c)[/tex]
Sau đó nó thành [tex]P=\frac{3}{2}t^2-3t=\frac{3}{2}(t-1)^2-\frac{3}{2}\geq -\frac{3}{2}[/tex]
Đúng rồi bạn, [tex]\sqrt{ab}\leq \frac{1}{2}(a+b);\, \sqrt[3]{abc}\leq \frac{1}{3}(a+b+c)[/tex]
Sau đó nó thành [tex]P=\frac{3}{2}t^2-3t=\frac{3}{2}(t-1)^2-\frac{3}{2}\geq -\frac{3}{2}[/tex]