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Fie \( a,b,c \in \mathfrak{R} ^+ \) astfel incat
\( a+b+c \geq \frac{1}{a}+ \frac{1}{b} + \frac{1}{c} \)
Aratati ca
\( a+b+c \geq \frac{3}{abc} \)

Off topic: Incepe sa imi placa LaTeX-ul

Din \( a+b+c\ \ge\ \frac 1a+\frac 1b+\frac 1c\ \Longrightarrow\ (a+b+c)^2\ \ge\ \left(\frac 1a+\frac 1b+\frac 1c\right)^2\ \stackrel{(\ast)}{\ge}\ 3\left(\frac 1{ab}+\frac 1{bc}+\frac 1{ca}\right)=\frac{3(a+b+c)}{abc} \)

Asadar, \( a+b+c\ \ge\ \frac 3{abc} \) . Am folosit inegalitatea cunoscuta : \( (x+y+z)^2\ \ge 3(xy+yz+zx)\ (\ast) \) .

bv, frumoasa solutia:)