A nice inequality. Because of the conditions, we can consider: $ x, y, z >0 $ real numbers as follows: $ \sqrt{ab-1}=z, \sqrt{bc-1}=x, \sqrt{ca-1}=y $ Now the inequality is "easier".

XbenX wrote: For $a,b,c$ real numbers not less than $1$ , prove that $$\frac{a+b+c}{4} \geq \frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}$$. Proof: $$\sum\frac{\sqrt{ab-1}}{b+c}\leq \frac{1}{2} \sum \sqrt{\frac{ab-1}{bc}}=\frac{1}{2} \sum \sqrt{(a-\frac{1}{b})\frac{1}{c}}\leq \frac{1}{4} \sum\left((a-\frac{1}{b})+\frac{1}{c}\right)=\frac{a+b+c}{4} .$$

Nice solution, dear professor Qing.

Lonesan wrote: Nice solution, dear professor Qing. Thank you very much. Bulgaria JBMO TST 2019: Let $a,b,c\geq 1.$ Find the maximum value of $\frac{\sqrt{a-1}}{a+b}+\frac{\sqrt{b-1}}{b+c}+\frac{\sqrt{c-1}}{c+a}.$

XbenX wrote: For $a,b,c$ real numbers not less than $1$ , prove that $$\frac{a+b+c}{4} \geq \frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}$$. Let $ a,b,c>0 $ and $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 $ . Prove that $$\frac{\sqrt{ab-1}}{a+b}+\frac{\sqrt{bc-1}}{b+c}+\frac{\sqrt{ca-1}}{c+a} \le \sqrt{2}$$

sqing wrote: XbenX wrote: For $a,b,c$ real numbers not less than $1$ , prove that $$\frac{a+b+c}{4} \geq \frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}$$. Let $ a,b,c>0 $ and $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 $ . Prove that $$\frac{\sqrt{ab-1}}{a+b}+\frac{\sqrt{bc-1}}{b+c}+\frac{\sqrt{ca-1}}{c+a} \le \sqrt{2}$$ So how can we solve it? Thank you.

sqing wrote: XbenX wrote: For $a,b,c$ real numbers not less than $1$ , prove that $$\frac{a+b+c}{4} \geq \frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}$$. Proof: $$\sum\frac{\sqrt{ab-1}}{b+c}\leq \frac{1}{2} \sum \sqrt{\frac{ab-1}{bc}}=\frac{1}{2} \sum \sqrt{(a-\frac{1}{b})\frac{1}{c}}\leq \frac{1}{4} \sum\left((a-\frac{1}{b})+\frac{1}{c}\right)=\frac{a+b+c}{4} .$$ Sorry, but how do you work out those somatories through inequalities and square roots? I’ve never seen something similar

For(n>2), $x_1,x_2,..,x_n$ real numbers not less than $1$ , prove that $$\frac{x_1+x_2+..+x_n}{4} \geq \frac{\sqrt{x_1x_2-1}}{x_2+x_3}+\frac{\sqrt{x_2x_3-1}}{x_3+x_4}+...+\frac{\sqrt{x_nx_1-1}}{x_1+x_2}$$

For $a_1,a_2,\cdots,a_n$$(n\geq 3)$ be real numbers not less than $1$ , prove that$$\frac{a_1+a_2+..+a_n}{4} \geq \frac{\sqrt{a_1a_2-1}}{a_2+a_3}+\frac{\sqrt{a_2a_3-1}}{a_3+a_4}+\cdots+\frac{\sqrt{a_{n-1}a_n-1}}{a_n+a_1}+\frac{\sqrt{a_na_1-1}}{a_1+a_2}$$See also here. The proof is exactly the same as the proof on #3.

Generalization: For(n>2), $x_1,x_2,..,x_n$ positive real numbers not less than $k>0$ , prove that $$ \frac{1}{4}\sum_{cyc}x_1+\frac{1-k^2}{4}\sum_{cyc}\frac{1}{x_1} \geq \sum_{cyc}\frac{\sqrt{x_1x_2-k^2}}{x_2+x_3}$$

AZOT1 wrote: Generalization: For(n>2), $x_1,x_2,..,x_n$ positive real numbers not less than $k>0$ , prove that $$ \frac{1}{4}\sum_{cyc}x_1+\frac{1-k^2}{4}\sum_{cyc}\frac{1}{x_1} \geq \sum_{cyc}\frac{\sqrt{x_1x_2-k^2}}{x_2+x_3}$$ solution we have \begin{align*}\sum_{cyc}\frac{\sqrt{x_1x_2-k^2}}{x_2+x_3} &\le \frac{1}{2}\sum_{cyc}\sqrt{\frac{x_1x_2-k^2}{x_2x_3}} &= \frac{1}{2}\sum_{cyc}\sqrt{\frac{1}{x_3}\left(x_1-\frac{k^2}{x_2}\right)} &\le \frac{1}{4}\sum_{cyc}\left(x_1-\frac{k^2}{x_2}+\frac{1}{x_3}\right) &= \frac{1}{4}\sum_{cyc}x_1+\frac{1-k^2}{4}\sum_{cyc}\frac{1}{x_1}\end{align*}

2019 ARMO Grade 9 P8 wrote: For $a,b,c$ be real numbers greater than $1$, prove that \[\frac{a+b+c}{4} \geqslant \frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}.\] In the beginning, obviously, it seemed ugly, but turns out it is actually very beautiful. Let $a=x+1$, $b=y+1$, $c=z+1$, notice that $x,y,z\geqslant 0$. Firstly, rewrite inequality as $$\frac{x+y+z+3}{2} \geqslant \frac{2\sqrt{xy+x+y}}{y+z+2}+\frac{2\sqrt{yz+y+z}}{x+z+2}+\frac{2\sqrt{xz+x+z}}{x+y+2}.$$ Applying AM-GM, we get $x+y+2=(x+1)+(y+1)\geqslant 2\sqrt{(x+1)(y+1)}\Longleftrightarrow \frac{1}{\sqrt{(x+1)(y+1)}}\geqslant \frac{2}{x+y+2}$. We do a similar thing for pairs $(x+1,z+1)$ and $(y+1,z+1)$ and obtain new inequality what we want to prove $$\frac{x+y+z+3}{2} \geqslant \sqrt{\frac{xy+x+y}{(y+1)(z+1)}}+\sqrt{\frac{yz+y+z}{(x+1)(z+1)}}+\sqrt{\frac{xz+x+z}{(x+1)(y+1)}}.$$Now let us apply Cauchy-Schwarz, $$\left(\sum_{cyc}\frac{xy+x+y}{y+1}\right)\left(\sum_{cyc} \frac{1}{x+1}\right)\geqslant \left(\sqrt{\frac{xy+x+y}{(y+1)(z+1)}}+\sqrt{\frac{yz+y+z}{(x+1)(z+1)}}+\sqrt{\frac{xz+x+z}{(x+1)(y+1)}}\right)^2.$$Notice that $$\sum_{cyc}\frac{xy+x+y}{y+1}=x+y+z+\sum_{cyc} \frac{x}{x+1}=x+y+z+3-\sum_{cyc} \frac{1}{x+1}.$$ Now let us set $t=x+y+z+3$ and $s=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$. Therefore, we need to show that $$\left(\frac{t}{2}\right)^2\geqslant (t-s)s\Longleftrightarrow t^2-4ts+4s^2\geqslant 0\Longleftrightarrow (t-2s)^2\geqslant 0.$$We are done.

$\sum_{\text{cyc}} \frac{\sqrt{ab-1}}{b+c} \stackrel{\text{AM-GM}}{\leq} \frac 12 \sum_{\text{cyc}} \sqrt{\frac{ab-1}{bc}}=\frac 12 \sum_{\text{cyc}} \sqrt{\frac 1c \left( a-\frac 1b \right) }\stackrel{\text{AM-GM}}{\leq} \frac 14 \sum_{\text{cyc}} \left( a+\frac 1c -\frac 1b \right)=\frac{a+b+c}{4}$ $\blacksquare$

For $a,b,c$ be real numbers greater than $1$, prove that \[\frac{a+b+c}{6} \geq \frac{\sqrt{ab-1}}{b+2c}+\frac{\sqrt{bc-1}}{c+2a}+\frac{\sqrt{ca-1}}{a+2b}\]

sqing wrote: For $a,b,c$ be real numbers greater than $1$, prove that \[\frac{a+b+c}{6} \geq \frac{\sqrt{ab-1}}{b+2c}+\frac{\sqrt{bc-1}}{c+2a}+\frac{\sqrt{ca-1}}{a+2b}\] Nice problem!