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Algebra Difficulty 6.6 National olympiad Prove it

14 Given ak⩾0,k=1,2,⋯ ,na_{k} \geqslant 0, k=1,2, \cdots, n. Define Ak=1k⋅∑i=1kaiA_{k}=\frac{1}{k} \cdot \sum_{i=1}^{k} a_{i}, prove:
∑k=1nAk2⩽4∑k=1nak2\sum_{k=1}^{n} A_{k}^{2} \leqslant 4 \sum_{k=1}^{n} a_{k}^{2}

Solution

14. If we set 1c∑k=1nAk2⩽∑k=1nAk⋅ak\frac{1}{c} \sum_{k=1}^{n} A_{k}^{2} \leqslant \sum_{k=1}^{n} A_{k} \cdot a_{k}, then we have ∑k=1nAk⋅ak⩽c⋅∑k=1nak2\sum_{k=1}^{n} A_{k} \cdot a_{k} \leqslant c \cdot \sum_{k=1}^{n} a_{k}^{2}. Thus, the problem can be transformed into an Abel method:
∑k=1nAkak=∑k=1nAk[kAk−(k−1)Ak−1]=∑k=1nkAk2−∑k=1n(k−1)AkAk−1⩾∑k=1nkAk2−12[∑k=1n(k−1)Ak2+∑k=1n(k−1)Ak−12]=12⋅∑k=1nAk2+12nAn2⩾12∑k=1nAk2\begin{aligned} \sum_{k=1}^{n} A_{k} a_{k} & =\sum_{k=1}^{n} A_{k}\left[k A_{k}-(k-1) A_{k-1}\right] \\ & =\sum_{k=1}^{n} k A_{k}^{2}-\sum_{k=1}^{n}(k-1) A_{k} A_{k-1} \\ & \geqslant \sum_{k=1}^{n} k A_{k}^{2}-\frac{1}{2}\left[\sum_{k=1}^{n}(k-1) A_{k}^{2}+\sum_{k=1}^{n}(k-1) A_{k-1}^{2}\right] \\ & =\frac{1}{2} \cdot \sum_{k=1}^{n} A_{k}^{2}+\frac{1}{2} n A_{n}^{2} \geqslant \frac{1}{2} \sum_{k=1}^{n} A_{k}^{2} \end{aligned}

Therefore,
∑k=1nAk2⩽4∑k=1nak2\sum_{k=1}^{n} A_{k}^{2} \leqslant 4 \sum_{k=1}^{n} a_{k}^{2}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.