Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.