Inegalitate cu variabile subunitare

[Claudiu Mindrila]

Sa se demonstreze ca, daca \( a,b,c\in [0,1] \) atunci: \( \frac{1}{5-ab}+\frac{1}{5-bc}+\frac{1}{5-ca} \ge \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{4} \).

[Marius Mainea]

Se arata ca

\( \frac{1}{5-bc}\ge \frac{\sqrt{b}+\sqrt{c}}{8} \)

\( \frac{1}{5-ca}\ge \frac{\sqrt{c}+\sqrt{a}}{8} \)

\( \frac{1}{5-ab}\ge \frac{\sqrt{a}+\sqrt{b}}{8} \) si apoi se aduna.


Intr-adevar \( (1-\sqrt{b})(1-\sqrt{c})\ge 0 \) deci \( \sqrt{b}+\sqrt{c}\le 1+\sqrt{bc} \)

si atunci \( \frac{\sqrt{b}+\sqrt{c}}{8}\le\frac{1+\sqrt{bc}}{8}\le \frac{1}{5-bc}\Longleftrightarrow (\sqrt{bc}-1)^2(\sqrt{bc}+3)\ge 0 \)

[Claudiu Mindrila]

A se vedea si aici .