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Problema. Determinati numerele reale pozitive \( a,b,c \) stiind ca indeplinesc simultan conditiile \( a^2+b^2+c^2=1 \) si \( \frac{b^2+c^2}{a}+\frac{a^2+c^2}{b}+\frac{a^2+b^2}{c} \leq 2\cdot\sqrt{3} \).
Petre Batranetu, Concursul "Cristian S. Calude", 2006

Vom arata ca in ipoteza \( a^2+b^2+c^2=1 \) rezulta \( \frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}\ge 2\sqrt{3} \)

Intr-adevar aplicand CBS \( LHS=\sum{\frac{a^2}{c}}+\sum{\frac{b^2}{c}}=\sum{\frac{a^4}{a^2c}}+\sum{\frac{b^4}{b^2c}}\ge\frac{(a^2+b^2+c^2)^2}{a^2c+b^2a+c^2b}+\frac{(a^2+b^2+c^2)^2}{b^2c+c^2a+a^2b}\ge \frac{4}{a^2(b+c)+b^2(c+a)+c^2(a+b)}\ge\frac{4}{\frac{1}{3}\cdot2(a+b+c)(a^2+b^2+c^2)}\ge 2\sqrt{3} \)

deoarece \( (a+b+c)^2\le 3(a^2+b^2+c^2) \) si sistemele \( (a^2,b^2,c^2) si (b+c,c+a,a+b) \) sunt invers orientate.

Asadar trebuie sa avem egalitate deci \( a=b=c=\frac{1}{\sqrt{3}} \)

Solutie. Fara a leza generalitatea putem presupune ca \( a=\min\left(a,b,c\right) \). Atunci \( \frac{1}{c}\leq\frac{1}{b}\leq\frac{1}{a} \) si \( a^{2}+b^{2}\leq a^{2}+c^{2}\leq b^{2}+c^{2} \). Cu inegalitatea lui Cebisev avem:
\( \frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a^{2}+b^{2}+c^{2}\right)\leq\sum\frac{a^{2}+b^{2}}{c} \) adica \( \sum\frac{a^{2}+b^{2}}{c}\geq\frac{2}{3}\left(\sum\frac{1}{a}\right) \). Dar conform inegalitatii C.B.S. avem: \( \frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq\frac{2}{3}\cdot9\cdot\frac{1}{a+b+c}\geq6\cdot\frac{1}{\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}}=\frac{6}{\sqrt{3}}=2\sqrt{3}. \)