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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Welcome to math.stackexchange! Did you try to take the squares? –  Davide Giraudo Jan 21 '12 at 21:54
Yes, I tried to take the squares. After cancellations, it becomes $\sqrt{s_{1}t_{1}}+\sqrt{t_{1}t_{2}}+\sqrt{s_{2}t_{2}}+\sqrt{s_{1}s_{2}}\ge\sqrt‌​{s_{1}t_{2}}+\sqrt{s_{2}t_{1}}+t_{1}$. Is this true for all $s_{1}\ge t_{1}\ge t_{2}\ge s_{2}\ge0$? –  user412 Jan 21 '12 at 22:28
I got the same; the main difficulty lies in comparing $\sqrt{s_1t_2}$ with $\sqrt{s_1s_2}+\sqrt{s_2t_2}$; perhaps find conditions for the former to be larger than the latter? –  Arturo Magidin Jan 21 '12 at 22:58
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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.