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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
Please choose more specific and descriptive titles for your questions. The purpose of the title is to summarize the question so as to represent it e.g. in the list of questions on the main page. To see why this title wasn't a good choice, imagine how that page would look if all users were to choose titles like this. – joriki Jan 22 '12 at 2:36

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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