It's here.

Amir.S wrote: Let $ x, y, z$ be real numbers greater than $ 1$ such that $ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = 2.$ Prove that $ \sqrt {x - 1} + \sqrt {y - 1} + \sqrt {z - 1}\leq \sqrt {x + y + z}.$ Substitute $ x = \frac {a + b + c}{b + c},y = \frac {a + b + c}{c + a},z = \frac {a + b + c}{a + b},$ where $ a,b,c > 0.$ Then $ \sum{\sqrt {x - 1}} = \sum{\sqrt{\frac {a}{b + c}}}\leq\sqrt {\sum{a}}\sqrt {\sum{\frac {1}{b + c}}} = \sqrt {\sum{x}},$ which is clearly true by Cauchy inequality. This problem is introduced as Example 2.2.8 on page 44 in new book : Ji Chen, Chao-Cheng Ji, Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture), Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

Amir.S wrote: Let $ x, y, z$ be real numbers greater than $ 1$ such that$ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = 2$ Prove that:$ \sqrt {x - 1} + \sqrt {y - 1} + \sqrt {z - 1}\leq \sqrt {x + y + z}$ $ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = 2$ $ \Longleftrightarrow \frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1,$ by Cauchy Inequality, we have $ \sqrt {x + y + z}$ $ =\sqrt {(x + y + z)\left(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}\right)}$ $ \ge \sqrt {x - 1} + \sqrt {y - 1} + \sqrt {z - 1}.$ done.

\[ 2=\sum\frac1x\implies1=\sum\frac{x-1}x\ge\frac{(\sum\sqrt{x-1})^2}{\sum x}\implies\sqrt{\sum x}\ge\sum\sqrt{x-1}\]

Johan Gunardi wrote: \[ 2 = \sum\frac1x\implies1 = \sum\frac {x - 1}x\ge\frac {(\sum\sqrt {x - 1})^2}{\sum x}\implies\sqrt {\sum x}\ge\sum\sqrt {x - 1}\] see the previous post you have posted the same exact solution

Sorry, I was typing my solution when the previous proof was posted.

Amir.S wrote: Let $x, y, z$ be real numbers greater than $1$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that \[\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\leq \sqrt{x+y+z}.\] Let $a_1,a_2,\cdots,a_n $ be real numbers greater than $1$ such that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=n-1 $$(n\ge 2).$ Prove that$$ \sqrt{a_1-1}+\sqrt{a_2-1}+\cdots+\sqrt{a_n-1}\leq \sqrt{a_1+a_2+\cdots+a_n}.$$

By C-S: $$(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2\leq (\sqrt{x(x-1)}^2+\sqrt{y(y-1)}^2+\sqrt{z(z-1)}^2)((\frac{1}{\sqrt{x}})^2+(\frac{1}{\sqrt{y}})^2+(\frac{1}{\sqrt{z}})^2=$$$$=(x^2 +y^2 +z^2 -x-y-z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2(x^2 +y^2 +z^2 -x-y-z)\leq (x+y+z)\leftrightarrow$$$$(x^2 +y^2 +z^2)\leq \frac{3}{2}(x+y+z) $$And using Chebyshev : $$\frac{(x^2 +y^2 +z^2)}{3}\frac{ (\frac {1}{x} + \frac {1}{y} + \frac {1}{z})}{3}\geq \frac{(x+y+z)}{3}\leftrightarrow (x^2 +y^2 +z^2)\leq \frac{3}{2}(x+y+z) $$

sqing wrote: Let $a_1,a_2,\cdots,a_n $ be real numbers greater than $1$ such that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=n-1 $$(n\ge 2).$ Prove that$$ \sqrt{a_1-1}+\sqrt{a_2-1}+\cdots+\sqrt{a_n-1}\leq \sqrt{a_1+a_2+\cdots+a_n}.$$

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Amir.S wrote: Let $x, y, z$ be real numbers greater than $1$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that \[\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\leq \sqrt{x+y+z}.\] $$(x+y+z)(\frac{x-1}{x} + \frac{y-1}{y} + \frac{z-1}{z} )\ge (\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$$ Let $ a,b,c >1 $ and $ \frac{1}{a}+\frac{2}{b}+\frac{1}{c} =2 . $ Prove that $$\sqrt{a+2b+c} \geq \frac{1}{ \sqrt{2}} \left(\sqrt{a-1}+ 2\sqrt{b-1} + \sqrt{c-1}\right) $$Let $ a,b,c \in (0,1) $ and $ab+bc+ca=4abc .$ Prove that $$\sqrt{a+2b+c}\geq \sqrt{\frac{2}{3}}\left(\sqrt{1-a}+\sqrt{2-b}+\sqrt{1-c}\right)$$Let $ a,b,c \in (0,1) $ and $ab+bc+ca=4abc .$ Prove that$$\sqrt{a+b+c}\geq \sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c}$$

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Generalization 2 Let $a_1,a_2,\cdots,a_n$ be positive real numbers greater than with order $\lambda_1,\lambda_2,\cdots,\lambda_n$ and $\sum_{i=1}^{n}{\dfrac{\lambda_i}{x_i}}=n-1$. Then prove that $$\sqrt{\sum_{cyc}{x_1}}\geq \sum_{j=1}^{n}{x_j-\lambda_j}$$

Let $x = a + 1$, $y = b + 1$, $z = c + 1$. Squaring the inequality and cancelling terms, we wish to prove $\frac{3}{2} \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$. Now $$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \implies 2abc + ab + bc + ca = 1 \implies 1 \ge 4\sqrt[4]{2abc}$$by AM-GM, so $\frac{1}{8} \ge abc \implies \frac{9}{8} \ge 3abc + ab + bc + ca$. Then $$\frac{9}{4} = \frac{9}{8} \left( \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \right) \ge \Bigl( bc(a + 1) + ca(b + 1) + ab(c + 1) \Bigl) \left( \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \right)$$but by Cauchy-Schwarz we also have $$\Bigl( bc(a + 1) + ca(b + 1) + ab(c + 1) \Bigl) \left( \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \right) \ge \left(\sqrt{bc} + \sqrt{ca} + \sqrt{ab} \right)^2$$so collectively we obtain $\frac{3}{2} \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$ as desired.