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$\sum_{j = 1}^{s} \sqrt{\alpha_j} \leq \sqrt{s} * \sqrt{\sum_{j = 1}^{s} \alpha_j}$

How would I go about proving this? I think Cauchy - Schwarz might be useful, but I can't quite get it to work.

Thanks!

Edit: $\alpha_j \geq 0$

Edit 2: There are $s$ values $\alpha_1 \dots \alpha_s$.

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    $\begingroup$ The Cauchy inequality would be $\sum c_j d_j \le (\sum c_j^2)^{1/2}(\sum d_j^2)^{1/2}$. Can you find values of $c_j, d_j$ to make this into yours? $\endgroup$
    – GEdgar
    Commented May 8, 2016 at 20:47
  • $\begingroup$ My answer to this question explains one possible approach. $\endgroup$
    – user228113
    Commented May 8, 2016 at 20:48

3 Answers 3

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You can directly apply Cauchy-Schwarz

$$ \sum_{j=1}^s \sqrt{\alpha_j} = \sum_{j=1}^s \sqrt{\alpha_j} \cdot 1 \leq \sqrt{\sum_{j=1}^s \left(\sqrt{\alpha_j}\right)^2} \cdot \sqrt{\sum_{j=1}^s 1^2} = \sqrt{\sum_{j=1}^s \alpha_j } \cdot \sqrt{s}.$$

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Use Jensen's inequality:

The function $\sqrt x$ is concave, hence $$\frac{\sum_{j=1}^s\sqrt\alpha_j}{s}\le\sqrt{\frac{\sum_{j=1}^s\alpha_j}{s}}. $$

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hint: $x_j = \dfrac{\alpha_j}{\displaystyle \sum_{j=1}^s \alpha_j}\implies \displaystyle \sum_{j=1}^s x_j = 1$. Thus you prove $\displaystyle \sum_{j=1}^s \sqrt{x_j} \leq \sqrt{s}$, and this is clearly true because $\sqrt{x_j} \leq 1$, $\forall j = \overline{1,s}$. Now apply Cauchy-Schwarz inequality: $\left(\displaystyle \sum_{j=1}^s \sqrt{x_j}\right)^2\leq s\cdot \displaystyle \sum_{j=1}^s x_j = s\implies $ the result follows.

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  • $\begingroup$ That makes a lot of sense. Thank you! $\endgroup$
    – PK5144
    Commented May 8, 2016 at 20:50

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