The following inequality is very nice and unsolved in AOPS for a long time.
I've tried to prove it without success. It's
For any positive $a,b,c$, prove that
$$\sqrt{\frac{2(a^2+b^2+c^2)}{ab+bc+ca}}+\sqrt{\frac{a+b}{c}\cdot\frac{b+c}{a}\cdot\frac{c+a}{b}}\ge \sqrt{\frac{a+b}{c}}+\sqrt{\frac{c+b}{a}}+\sqrt{\frac{a+c}{b}}.$$
See also AOPS
As a user pointed out in the link, the problem has a geometric equivalent form.
$$\sum_{cyc}\sin\frac A2 \le 1 + \sqrt{\frac {s^2 – 8Rr – 2r^2}{2R(4R + r)}}.$$
I'd like to prove it by algebra way.
Since equality holds at $a=b=c$ and $a^2+b^2+c^2\ge ab+bc+ca$ we can consider the stronger one$$\sqrt{2}+\sqrt{\frac{a+b}{c}\cdot\frac{b+c}{a}\cdot\frac{c+a}{b}}\ge \sqrt{\frac{a+b}{c}}+\sqrt{\frac{c+b}{a}}+\sqrt{\frac{a+c}{b}}.$$
Now let $$x=\sqrt{\frac{a+b}{c}};y=\sqrt{\frac{c+b}{a}};z=\sqrt{\frac{a+c}{b}}$$then $x,y,z>0$ and $xyz\ge 2\sqrt{2}.$
We'll prove $$\sqrt{2}+xyz\ge x+y+z.$$
I got stuck here.
It would be great if you help me prove or disprove my subsequent inequality. Now, I see it is wrong inequality.
Also, you can share any ideas and comments. Thank you very much.
Thank you @RiverLi. The current proof is nice enough but it is not easy. Is there a nice proof?
Best Answer
I think the stronger one is not right. For example, let $a=8$, $b=1$, $c=1$, then
$$ \sqrt 2 + \sqrt{\frac{a+b}{c}\cdot \frac{b+c}{a} \cdot \frac{c+a}{b}}=\sqrt2 + \frac{9}{2}<6<\frac{13}{2}={\sqrt{\frac{a+b}{c}}+ \sqrt{\frac{b+c}{a}}+ \sqrt{\frac{c+a}{b}}} $$ PS: I don't have enough reputation to add a comment under your post, so I directly comment here.