[tex] f(x)=(a+b+c)x^2 -6x+\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} [/tex]
Graficul funcției nu taie Ox, asta înseamnă ca f(x)=0 nu are soluții reale. Cu alte cuvinte, [tex] \Delta <0 [/tex]
[tex] \Rightarrow b^2 -4ac <0 \\ \Rightarrow 36-4(a+b+c)(\tfrac{1}{a} + \tfrac{1}{b} + \tfrac{1}{c} ) <0 \\ (a+b+c)(\tfrac{1}{a} + \tfrac{1}{b}+ \tfrac{1}{c} ) >9 \\ 1+\tfrac{a}{b}+\tfrac{a}{c} + \tfrac{b}{a} + 1+\tfrac{b}{c} + \tfrac{c}{a}+\tfrac{c}{b}+1 >9 \\ ( \underbrace{\tfrac{a}{b} + \tfrac{b}{a}}_{\ge 2} )+ (\underbrace{\tfrac{b}{c} + \tfrac{c}{b}}_{\ge 2} )+ (\underbrace{\tfrac{a}{c} + \tfrac{c}{a}}_{\ge 2}) >6 \\ \Rightarrow \tt Adevarat , \ \forall a,b,c \in \mathbb{R} _{+} [/tex]
