Áp dụng BĐT Cauchy ta có: $\frac{y+z}{2x}+\frac{y+z}{2x}+1\ge3\sqrt[3]{\left(\frac{y+z}{2x}\right)^2}$$\Leftrightarrow \frac{x+y+z}{x}\ge 3\sqrt[3]{\left(\frac{y+z}{2x}\right)^2}$$\Leftrightarrow \sqrt[3]{\left(\frac{x}{y+z}\right)^2}\ge\frac{3}{\sqrt[3]{4}}.\frac{x}{x+y+z}$Tương tự: $\sqrt[3]{\left(\frac{y}{x+z}\right)^2}\ge\frac{3}{\sqrt[3]{4}}.\frac{y}{x+y+z};\sqrt[3]{\left(\frac{z}{x+y}\right)^2}\ge\frac{3}{\sqrt[3]{4}}.\frac{z}{x+y+z}$Cộng 3 BĐT trên lại ta có đpcm.Dấu bằng xảy ra khi: $x=y=z$
Áp dụng BĐT Cauchy ta có: $\dfrac{y+z}{2x}+\dfrac{y+z}{2x}+1\ge3\sqrt[3]{\left(\dfrac{y+z}{2x}\right)^2}$$\Leftrightarrow \dfrac{x+y+z}{x}\ge3\sqrt[3]{\left(\dfrac{y+z}{2x}\right)^2}$$\Leftrightarrow \sqrt[3]{\left(\dfrac{x}{y+z}\right)^2}\ge\dfrac{3}{\sqrt[3]{4}}.\dfrac{x}{x+y+z}$Tương tự: $\sqrt[3]{\left(\dfrac{y}{x+z}\right)^2}\ge\dfrac{3}{\sqrt[3]{4}}.\dfrac{y}{x+y+z};\sqrt[3]{\left(\dfrac{z}{x+y}\right)^2}\ge\dfrac{3}{\sqrt[3]{4}}.\dfrac{z}{x+y+z}$Cộng 3 BĐT trên lại ta có đpcm.Dấu bằng xảy ra khi: $x=y=z$
Áp dụng BĐT Cauchy ta có: $\frac{y+z}{2x}+\frac{y+z}{2x}+1\ge3\sqrt[3]{\left(\frac{y+z}{2x}\right)^2}$$\Leftrightarrow \frac{x+y+z}{x}\ge
3\sqrt[3]{\left(\frac{y+z}{2x}\right)^2}$$\Leftrightarrow \sqrt[3]{\left(\frac{x}{y+z}\right)^2}\ge\frac{3}{\sqrt[3]{4}}.\frac{x}{x+y+z}$Tương tự: $\sqrt[3]{\left(\frac{y}{x+z}\right)^2}\ge\frac{3}{\sqrt[3]{4}}.\frac{y}{x+y+z};\sqrt[3]{\left(\frac{z}{x+y}\right)^2}\ge\frac{3}{\sqrt[3]{4}}.\frac{z}{x+y+z}$Cộng 3 BĐT trên lại ta có đpcm.Dấu bằng xảy ra khi: $x=y=z$