Maybe this can help, but I can't finish.. The condition $ \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2$ is equivalent to $abc= a+b+c+2$ and therefore we can make the substitution $ a=\frac{y+z}{x}, b=\frac{x+z}{y}$ and $c=\frac{x+y}{z} $ Hence the inequality becomes \[{ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right) \] Any thoughts on this?

I think now you can use Cebisev using $ y+z-x $ is decreasing in $ x $ and $ x(y+z) $ is increasing in $ x $.

hatchguy wrote: \[{ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right) \] $\Leftrightarrow \sum \left ( \sqrt{\frac{y+z}{2x}} - \sqrt{\frac{2x}{y+z}} \right ) \geq 0$ $\Leftrightarrow \sum \left ( \frac{y+z-2x}{\sqrt{x(y+z)}} \right ) \geq 0 \Leftrightarrow \sum \left ( \frac{y-x}{\sqrt{x(y+z)}} + \frac{x-y}{\sqrt{y(x+z)}} \right ) \geq 0 $ $= \sum \left ( (x-y)(\frac{1}{\sqrt{y(z+x)}} - \frac{1}{\sqrt{x(y+z)}}) \right ) $ $= \sum \left ( \frac{z(x-y)^2}{ \left [\sqrt{y(z+x)}+\sqrt{x(y+z)} \right ] \sqrt{xy(z+x)(z+y)} } \right ) \geq 0$

Nice (end of )solution Mahanmath!.

amparvardi wrote: Let $a, b, c$ be positive real numbers such that \[\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.\] Prove that \[\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.\] This is solution that I found during the competition:

hatchguy wrote: Hence the inequality becomes \[{ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right) \] Any thoughts on this? Substitute $r=\sqrt{x}, s=\sqrt{y}, t=\sqrt{z}$ and use $\sqrt{r^2+s^2}\ge \frac1 {\sqrt{2}}(r+s)$ in the numerators as well as the denominators to get: \[ \sum \frac{r+s}t \ge 4 \sum \frac r {s+t}\] This inequality has been discussed before and is very easy .

why are you hiting your head on the desk ? \[ \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2. \] \[ \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1. \] so $ a=\frac{y+z}{x}, b=\frac{x+z}{y} $and $c=\frac{x+y}{z}$,

hatchguy wrote: \[{ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right) \] Generalisation: If $x_1,x_2,\cdots,x_n>0,S=x_1+x_2+\cdots+x_n,n\ge 2$,then \[\sqrt{\frac{S-x_1}{x_1}}+\sqrt{\frac{S-x_2}{x_2}}+\cdots+\sqrt{\frac{S-x_n}{x_n}} \ge \] \[ \ge (n-1) \left(\sqrt{\frac{x_1}{S-x_1}}+\sqrt{\frac{x_2}{S-x_2}}+\cdots+\sqrt{\frac{x_n}{S-x_n}}\right)\] Proof,notice that (1)\[{\frac{S-x_1}{x_1}+\frac{S-x_2}{x_2}+\cdots+\frac{S-x_n}{x_n}}\ge\] \[\ge (n-1)^2 \left(\frac{x_1}{S-x_1}+\frac{x_2}{S-x_2}+\cdots+\frac{x_n}{S-x_n}\right)\] This is because,By cauchy inequality, \[(n-1)^2\frac{x_1}{S-x_1}\le \frac{x_1}{x_2}+\frac{x_1}{x_3}+\cdots+\frac{x_1}{x_n}\] And the other n-1 similar inequalies,then sum them up. (2)For $1\le i<j\le n$,we have: \[\sqrt{\frac{(S-x_i)(S-x_j)}{x_ix_j}}\ge 1+\sum_{k=1,k \neq i,j}\frac{x_k}{\sqrt{x_ix_j}} \] (3)For $1\le i<j\le n$,we have: \[\sqrt{\frac{x_ix_j}{(S-x_i)(S-x_j)}}\le \frac{1}{2}\left(\frac{x_i}{S-x_j}+\frac{x_j}{S-x_i}\right)\] By,(1),(2),(3).We can square the orginal inequality,then not hard to conclude a proof.

2011年中欧数学奥林匹克不等式试题的一般结论

Georg-A. wrote: hatchguy wrote: Hence the inequality becomes \[{ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right) \] Any thoughts on this? Substitute $r=\sqrt{x}, s=\sqrt{y}, t=\sqrt{z}$ and use $\sqrt{r^2+s^2}\ge \frac1 {\sqrt{2}}(r+s)$ in the numerators as well as the denominators to get: \[ \sum \frac{r+s}t \ge 4 \sum \frac r {s+t}\] This inequality has been discussed before and is very easy . The inequality \[ \sum \frac{r+s}t \ge 4 \sum \frac r {s+t}\] not equivalent \[{ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right) \]

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=114791&hilit=inequality http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2869656

Amir Hossein wrote: Let $a, b, c$ be positive real numbers such that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$ Prove that \[\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.\] Let $a, b, c$ be positive real numbers such that $ \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$ Prove that $$a^2 +b^2+c^2 \geq 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).$$$$abc+4\geq 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).$$

For three positive real numbers a, b, c, prove that $$ \sqrt{\frac{a+b}{c}} +\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}{ \geq } 2(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}})$$https://artofproblemsolving.com/community/c6h397101p2208537 https://artofproblemsolving.com/community/c6h475161p2661111 https://artofproblemsolving.com/community/c6h511112p2869656 Amir Hossein wrote: Let $a, b, c$ be positive real numbers such that \[\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.\]Prove that \[\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.\] $$cosA+cosB+cosC\geq 2(cosBcosC+cosCcosA+cosAcosB)$$