Proof that this double summation is nonnegative

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I need to prove this equation

$$\sum_{i=1}^K {\bigg(x_i\bigg(x_is_i-\sum_{j=1}^K {x_js_is_j}\bigg)\bigg)}\ge0$$

knowing that

  • $K\in\mathbb{N},K\ge2$
  • $x,s\in\mathbb{R}^K$
  • $\forall {k\in\{1,\dots,K\}}:s_k\in]0,1[$
  • $\sum_{k=1}^K {s_k}=1$

I tried to develop the first formula but I can't find a solution $$\sum_{i=1}^K {\bigg(x_i\bigg(x_is_i-\sum_{j=1}^K {x_js_is_j}\bigg)\bigg)}$$ $$=\sum_{i=1}^K {\bigg(x_i^2s_i-x_is_i\sum_{j=1}^K {x_js_j}\bigg)}$$ $$=\sum_{i=1}^K {x_i^2s_i}-\sum_{i=1}^K {\sum_{j=1}^K {x_ix_js_is_j}}$$ $$=\sum_{i=1}^K {x_i^2s_i}-\bigg(\sum_{i=1}^K {x_is_i}\bigg)^2$$

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$\bigg(\sum_{i=1}^K {x_is_i}\bigg)^2=\bigg(\sum_{i=1}^K {(x_i\sqrt {s_i}) (\sqrt {s_i})\bigg)}^2\leq (\sum_{i=1}^K {(x_i\sqrt {s_i})^{2} })$ by Cauchy-Schwarz inequality.