Yes, this one has nothing to do with geometry... At first, let $b_{i}=s-a_{i}$. Then, $b_{2}+b_{3}+b_{4}=3s-\left( a_{2}+a_{3}+a_{4}\right) =3s-\left( 2s-a_{1}\right) =a_{1}+s$, and similarly $b_{3}+b_{4}+b_{1}=a_{2}+s$, $b_{4}+b_{1}+b_{2}=a_{3}+s$ and $b_{1}+b_{2}+b_{3}=a_{4}+s$, so that our inequality becomes \[ \frac{1}{b_{2}+b_{3}+b_{4}}+\frac{1}{b_{3}+b_{4}+b_{1}}+\frac{1}{b_{4}+b_{1}+b_{2}}+\frac{1}{b_{1}+b_{2}+b_{3}} \\ \leq \frac{2}{9}\left( \frac{1}{\sqrt{b_{1}b_{2}}}+\frac{1}{\sqrt{b_{1}b_{3}}}+\frac{1}{\sqrt{b_{1}b_{4}}}+\frac{1}{\sqrt{b_{2}b_{3}}}+\frac{1}{\sqrt{b_{2}b_{4}}}+\frac{1}{\sqrt{b_{3}b_{4}}}\right). \] Now, let \[ c_{i}=\frac{1}{b_{i}} \] Then, our inequality can be rewritten as \[ \frac{1}{\frac{1}{c_{2}}+\frac{1}{c_{3}}+\frac{1}{c_{4}}}+...\leq \frac{2}{9}\left( \sqrt{c_{1}c_{2}}+\sqrt{c_{1}c_{3}}+...+\sqrt{c_{3}c_{4}}\right), \] or \[ \frac{3}{\frac{1}{c_{2}}+\frac{1}{c_{3}}+\frac{1}{c_{4}}}+...\leq \frac{2}{3}\left( \sqrt{c_{1}c_{2}}+\sqrt{c_{1}c_{3}}+...+\sqrt{c_{3}c_{4}}\right). \] The left member consists of four harmonic means; as any harmonic mean is less or equal to the corresponding geometric mean, we have \[ \frac{3}{\frac{1}{c_{2}}+\frac{1}{c_{3}}+\frac{1}{c_{4}}}+...\leq \sqrt[3]{c_{2}c_{3}c_{4}}+... \] In the following, we will prove \[ \sqrt[3]{c_{2}c_{3}c_{4}}+...\leq \frac{2}{3}\left( \sqrt{c_{1}c_{2}}+\sqrt{c_{1}c_{3}}+...+\sqrt{c_{3}c_{4}}\right); \] once this inequality will be verified, the problem will be solved. We make another substitution (the last one): $d_{i}=\sqrt[6]{c_{i}}$. Then our inequality becomes \[ d_{2}^{2}d_{3}^{2}d_{4}^{2}+...\leq \frac{2}{3}\left( d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}\right),\] or \[ \frac{d_{2}^{2}d_{3}^{2}d_{4}^{2}+...}{4}\leq \frac{d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}}{6}. \] But the Great Cursed Symmetric Means inequality (BTW, it's the first time I'm applying it in my life) yields \[ \sqrt[3]{\frac{d_{2}^{2}d_{3}^{2}d_{4}^{2}+...}{4}}\leq \sqrt{\frac{d_{1}^{2}d_{2}^{2}+d_{1}^{2}d_{3}^{2}+...+d_{3}^{2}d_{4}^{2}}{6}}, \] while the Power Mean inequality tells us \[ \sqrt{\frac{d_{1}^{2}d_{2}^{2}+d_{1}^{2}d_{3}^{2}+...+d_{3}^{2}d_{4}^{2}}{6}}\leq \sqrt[3]{\frac{d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}}{6}}, \] so that we have \[ \sqrt[3]{\frac{d_{2}^{2}d_{3}^{2}d_{4}^{2}+...}{4}}\leq \sqrt[3]{\frac{d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}}{6}},\] and \[ \frac{d_{2}^{2}d_{3}^{2}d_{4}^{2}+...}{4}\leq \frac{d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}}{6}. \] Proof complete! I hope I have not applied some inequalities with wrong sign again Darij [Admin Edit: PLEASE do not use the \Huge and [ tex ] tags any more ... ]

My solution is the following: we have \[\frac{2}{9}\sum_{1 \leq i < j \leq 4}\frac{1}{\sqrt{(s - a_i)(s - a_j)}} \geq \frac{2}{9}2\sum_{1 \leq i < j \leq 4}\frac{1}{(s - a_i) + (s - a_j)} = \frac{4}{9}\sum_{1 \leq i < j \leq 4}\frac{1}{a_i + a_j}.\] Let's denote $a_1 = a$, $a_2 = b$, $a_3 = c$ and $a_4 = d$. We should prove now that \[\frac{2}{9}\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{a + d} + \frac{1}{b + c} + \frac{1}{b + d} + \frac{1}{c + d}\right) \geq\]\[\frac{1}{3a + b + c + d} + \frac{1}{a + 3b + c + d} + \frac{1}{a + b + 3c + d} + \frac{1}{a + b + c + 3d}.\] By Cauchy we have: \[\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{a + d}\right)\left((a + b) + (a + c) + (a + d)\right) \geq 9\]\[\Rightarrow \frac{1}{9}\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{a + d}\right) \geq \frac{1}{3a + b + c + d}.\] Adding the three analogous inequalities to this one we get the result.

darij grinberg wrote: so that we have \[ \sqrt[3]{\frac{d_{2}^{2}d_{3}^{2}d_{4}^{2}+...}{4}}\leq \sqrt[3]{\frac{d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}}{6},\] and If you could add another } after {6} you will get the image you want and all the error messages will go! Putting in the missing } gives \[ \sqrt[3]{\frac{d_{2}^{2}d_{3}^{2}d_{4}^{2}+...}{4}}\leq \sqrt[3]{\frac{d_{1}^{3}d_{2}^{3}+d_{1}^{3}d_{3}^{3}+...+d_{3}^{3}d_{4}^{3}}{6}},\]

Arne: yeah, your solution is a piece of cake compared to mine, thanks Stevem: Thanks a lot for pointing out the error in my LaTeX code. Actually, please excuse me but I have NO CLUE about what I am doing when I write LaTeX tags. I have never worked with LaTeX, instead I have always been using Scientific WorkPlace; this is a program which enables you to write TeX files without really knowing TeX, just by a user-friendly WYSIWYG interface. Hence, when I need a formula, I just type it in this interface and copy&paste it into the forum. If I try to edit it manually afterwards, I usually get lots of errors. Sorry again. I used \Huge in order to have the fractions large enough such that they can be read properly. But I see it's not really necessary Darij

No need to apologise. I too got into LaTeX using Scientific Workplace, which is a great program for typing maths since it's so natural and you don't need to use the mouse a lot. But on occasions it uses its own system which isn't recognised elsewhere, which makes it awkward. So it's really good to learn LaTeX and it will be extremely useful to you in the future. Some university mathematics departments have special classes in LaTeX to help their students. Because it's so easy to make mistakes I use Preview a lot, so it looks great when posted - no-one sees the hundreds of errors on the way!