If $s_{1}\ge t_{1}\ge t_{2}\ge s_{2}\ge0$, does one always have $(s_{1}-t_{1}+s_{2}+t_{2})^{1/2}\ge\sqrt{s_{1}}-\sqrt{t_{1}}+\sqrt{t_{2}}-\sqrt{s_{2}}$? Thanks a lot!
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Both sides of the inequality are positive, so we can square it. We have to show that $$ \begin{multline} I=\frac12[s_1-t_1+s_2 + t_2 - (\sqrt{s_1}-\sqrt{t_1} + \sqrt{t_2} - \sqrt{s_2})^2]\\ = \sqrt{s_{1}t_{1}}+\sqrt{t_{1}t_{2}}+\sqrt{s_{2}t_{2}}+\sqrt{s_{1}s_{2}}-\sqrt{s_{1}t_{2}}-\sqrt{s_{2}t_{1}}-t_{1} \ge 0. \end{multline}$$ We want to show that decreasing $s_2$ makes the left-hand side always smaller. To this end, we calculate the derivative of $I$ with respect to $s_2$: $$ \frac{\partial I}{\partial s_2} = \frac{1}{2\sqrt{s_2}} \underbrace{(\sqrt{s_1} - \sqrt{t_1} + \sqrt{t_2})}_{\geq\sqrt{s_2}}\geq \frac12.$$ Thus the inequality is tightest when $s_2=0$. Setting $s_2=0$ yields $$I|_{s_2=0}= \sqrt{s_{1}t_{1}}+\sqrt{t_{1}t_{2}}-\sqrt{s_{1}t_{2}}-t_{1} = (\sqrt{s_1} - \sqrt{t_1}) (\sqrt{t_1} -\sqrt{t_2}) \geq 0$$ and thus the original inequality is always fulfilled. |
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