Folosim relația: [tex]\bf \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a+\dfrac{1}{a}\geq2,\ \ \ a>0[/tex]
[tex]\it \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+x+y+z =\Big(\underbrace{\dfrac{1}{x}+x}_{\geq2}\Big)+\Big(\underbrace{\dfrac{1}{y}+y}_{\geq2}\Big)+\Big(\underbrace{\dfrac{1}{z}+z}_{\geq2}\Big) \geq6 \Rightarrow\\ \\ \\ \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+x+y+z\geq6\Rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+1 \geq6|_{-1}\Rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\geq5[/tex]
Deci, minimul expresiei din enunț este egal cu 5.
