Dacă  x_k \in \mathbb R şi  a_k > 0 , atunci:

 \frac {x_1^2}{a_1} + \frac {x_2^2}{a_2} + \cdots + \frac {x_n^2}{a_n} \ge \frac {(x_1 + x_2 + \cdots + x_n)^2}{a_1 + a_2 + \cdots + a_n},   (inegalitatea lui Bergström)

cu egalitate dacă şi numai dacă:

 \frac {x_1}{a_1} =  \frac {x_2}{a_2} = \cdots =  \frac {x_n}{a_n}.

Generalizarea inegalităţii lui Bergström este dată de următoarea teoremă:


Dacă  x_k \in \mathbb R şi  a_k > 0, \; k \in \{ 1, 2, \cdots , n \}, atunci:

 \frac {x_1^2} {a_1} + \frac {x_2^2} {a_2} + \cdots + \frac {x_n^2} {a_n}  \ge \frac {(x_1 + x_2 + \cdots + x_n)^2}{a_1 +a_2 + \cdots + a_n} + \max_{1 \le j < \le n} \frac {(a_i x_j - a_j x_i)^2}{a_i a_j (a_i + a_j)}.

Resurse Edit

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