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a) Fie \( a, b, c \) numere reale astfel incat \( a\in [0;1], b\in [0; \frac{1}{2}], c\in [0;\frac{1}{3}] \) si \( 2ab+6bc+3ac=1 \).
Demonstrati ca \( a^2+2b^2+3c^2\le 2 \).
Prof. univ. Dumitru Acu, Sibiu
b) Sa se demonstreze ca:
\( 1-\frac{2^2+3}{2^4-2}-\frac{4^2+5}{4^4-5}-.....-\frac{2010^2+2011}{2010^4-2010}=\frac{1}{2010}+\frac{3^2+4}{3^4-3}+\frac{5^2+6}{5^4-5}+.....+\frac{2009^2+2010}{2009^4-2009} \).
Prof. Doina Firicel, Calafat

Sunt intr-o dilema la punctul a) :

Daca : \( a \in [0;1] \Longleftrightarrow 0 \le a \le 1 \Longleftrightarrow 0 \le a^2 \le 1 \)

\( b \in [0;\frac{1}{2}] \Longleftrightarrow 0 \le b \le \frac{1}{2} \Longleftrightarrow 0 \le b^2 \le \frac{1}{4} \Longleftrightarrow 0 \le 2b^2 \le \frac{1}{2} \)

\( c \in [0;\frac{1}{3}] \Longleftrightarrow 0 \le c \le \frac{1}{3} \Longleftrightarrow 0 \le c^2 \le \frac{1}{9} \Longleftrightarrow 0 \le 3c^2 \le \frac{1}{3} \)

Prin insumare : \( a^2+2b^2+3c^2 \le 1+\frac{1}{2}+\frac{1}{3}=1+\frac{5}{6} \lt 2 \)

Cred ca enuntul este gresit, altfel conditia \( 2ab+6bc+3ac=1 \) n-ar avea niciun rost.