Toán 8 Chứng minh bất đẳng thức [tex]A\leq 1[/tex]

Ta có:[tex]\frac{1}{x+y}+\frac{1}{x+z}\geq \frac{4}{2x+y+z}\Rightarrow \frac{1}{2x+y+z}\leq \frac{1}{4(x+y)}+\frac{1}{4(x+z)}[/tex]
Lại có:[tex]\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y};\frac{1}{x}+\frac{1}{z}\geq \frac{4}{x+z}\Rightarrow \frac{1}{4(x+y)}+\frac{1}{4(x+z)}\leq \frac{1}{16}(\frac{2}{x}+\frac{1}{y}+\frac{1}{z})[/tex][tex]\Rightarrow \frac{1}{2x+y+z}\leq \frac{1}{16}(\frac{2}{x}+\frac{1}{y}+\frac{1}{z})[/tex]
Tương tự, ta cũng có: [tex]\frac{1}{x+2y+z}\leq \frac{1}{16}(\frac{1}{x}+\frac{2}{y}+\frac{1}{z});\frac{1}{x+y+2z}\leq \frac{1}{16}(\frac{1}{x}+\frac{1}{y}+\frac{2}{z})[/tex]
[tex]\Rightarrow A\leq \frac{1}{16}(\frac{4}{x}+\frac{4}{y}+\frac{4}{z})=\frac{1}{4}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=1[/tex]
Dấu "=" xảy ra khi [tex]x=y=z=\frac{3}{4}[/tex]