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Daca a,b,c>0 cu a+b+c=1 , atunci

\( \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}\ge \frac{9}{10}. \)

Marin Chirciu , RMT/4/2008

\( \sum {\frac {a}{bc+1}} \) \( = \) \( \sum {\frac {a^2}{abc+a} \) \( \frac {Cauchy}{\geq} \) \( \frac {1}{3abc+1} \)
\( \sum {\frac {a}{bc+1}} \geq {\frac {9}{10} \Leftrightarrow \frac {1}{3abc+1} \geq \frac {9}{10} \Leftrightarrow \frac {1}{27} \geq abc \) \( \Leftrightarrow \) \( \frac {a+b+c}{3} \geq \sqrt[3] {abc} \) .

Solutie.

Prelucram prima fractie, \( \frac{a}{bc+1}=\frac{a\left(bc+1\right)-abc}{bc+1}=a-\frac{abc}{bc+1} \). Prin sumare rezulta ca
\( LHS=\sum a- \sum \frac{abc}{bc+1}=1-abc(\sum \frac{1}{bc+1}) \).
Insa cum \( \sqrt[3]{abc} \leq \frac{a+b+c}{3} \Longrightarrow -abc \geq \frac{1}{27}.(1) \)
Apoi, din inegalitatea Cauchy-Buniakowski-Schwarz avem \( \sum \frac{1}{bc+1} \geq \frac{9}{ab+bc+ca+3} \geq \frac{9}{\frac{(a+b+c)^2}{3}+3}=\frac{27}{10}.(2) \)
Inmultind inegalitatile \( (1) \) si \( (2) \) obtinem ca \( \sum \frac{abc}{bc+1} \geq -\frac{1}{10} \) si deci \( LHS \geq 1-\frac{1}{10}=\frac{9}{10} \)

Pentru alte 2 solutii a se vedea "C. Mindrila- In legatura cu problema O.VII.252, R.M.T. 1/2009"