a,b ≥ 0
[tex] (a+b+1)(a+b+ab) = ab\Big( \dfrac{1}{a}+\dfrac{1}{b} +\dfrac{1}{ab}\Big)(ab+a+b)\\ \\ \text{Aplicam inegalitatea mediilor:}\\ \\ \dfrac{a+b+ab}{3} \geq \sqrt[3]{a\cdot b\cdot ab} \\ \\ \text{Si similar:} \\ \\ \dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}}{3} \geq \sqrt[3]{\dfrac{1}{a\cdot a\cdot ab}}[/tex]
[tex] \text{Inmultim inegalitatile:} \\ \\ \dfrac{a+b+ab}{3}\cdot \dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}}{3}\geq \sqrt[3]{a\cdot b\cdot ab}\cdot \sqrt[3]{\dfrac{1}{a\cdot a\cdot ab}} \\ \\ \dfrac{ \Big(\dfrac{1}{a}+\dfrac{1}{b} +\dfrac{1}{ab}\Big)(a+b+ab)}{9} \geq 1 \Big|\cdot 9ab \\ \\ \Rightarrow ab\Big(\dfrac{1}{a}+\dfrac{1}{b} +\dfrac{1}{ab}\Big)(a+b+ab) \geq 9ab \\ \\ \Rightarrow (a+b+1)(a+b+ab) \geq 9ab [/tex]
