Đáp án:$ E\ge \dfrac{3}{2}$
Giải thích các bước giải:
$\dfrac{1}{x^3.(y+z)}=\dfrac{\dfrac{1}{x^2}}{x(y+z)}=\dfrac{(\dfrac{1}{x})^2}{\dfrac{1}{y}+\dfrac{1}{z}}$
$\dfrac{1}{y^3.(x+z)}=\dfrac{\dfrac{1}{y^2}}{y(x+z)}=\dfrac{(\dfrac{1}{y})^2}{\dfrac{1}{x}+\dfrac{1}{z}}$
$\dfrac{1}{z^3.(x+y)}=\dfrac{\dfrac{1}{z^2}}{z(x+y)}=\dfrac{(\dfrac{1}{z})^2}{\dfrac{1}{x}+\dfrac{1}{y}}$
$\rightarrow E=\dfrac{(\dfrac{1}{x})^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{(\dfrac{1}{y})^2}{\dfrac{1}{x}+\dfrac{1}{z}}+\dfrac{(\dfrac{1}{z})^2}{\dfrac{1}{x}+\dfrac{1}{y}}$
$\rightarrow E\ge \dfrac{(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})^2}{\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}}$
$\rightarrow E\ge \dfrac{(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})^2}{2(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})}$
$\rightarrow E\ge \dfrac{1}{2}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})\ge \dfrac{1}{2}.3\sqrt[3]{\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}}$
$\rightarrow E\ge \dfrac{3}{2}$
