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$\left(\sum_i a_i\right)^2\geq \sum_i \left(a_i^2+2\sum_{j>i} a_ia_j\right)\geq \sum_i a_i^2+2\sum_i\sum_{j>i} a_j^2$ But that last sum is just $a_1^2+3a_2^2+\ldots+(2n-1)a_n^2$, so we're done.

shobber wrote: If $ a_1 \geq a_2 \geq \cdots \geq a_n \geq 0$ and $ a_1 + a_2 + \cdots + a_n = 1$, then prove: \[ a_1^2 + 3a_2^2 + 5a_3^2 + \cdots + (2n - 1)a_n^2 \leq 1 \] Only Karamata:)

Arne wrote: Hint:

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Can you please suggest a solution

$$LHS = \sum_{i=1}^n a_i^2 + 2\sum_{i=1}^{n-1} ia_{i+1}^2 \le \sum_{i=1}^n a_i^2 + 2\sum_{i=1}^{n-1} a_{i+1}(a_1+a_2+...+a_i) = \left(\sum_{i=1}^n a_i\right)^2 = 1$$

blahblahblah wrote: $\left(\sum_i a_i\right)^2\geq \sum_i \left(a_i^2+2\sum_{j>i} a_ia_j\right)\geq \sum_i a_i^2+2\sum_i\sum_{j>i} a_j^2$ But that last sum is just $a_1^2+3a_2^2+\ldots+(2n-1)a_n^2$, so we're done. Please, can someone explain the last inequality?