parmenides51 wrote: Let $x,y,z\in \mathbb{R}^+_0$ such that $xy+yz+zx=1$. Prove that $$\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}\ge 2+\frac{1}{\sqrt{2}}.$$ ( Anonymous314) Very known inequality.

parmenides51 wrote: Let $x,y,z\in \mathbb{R}^+_0$ such that $xy+yz+zx=1$. Prove that $$\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}\ge 2+\frac{1}{\sqrt{2}}.$$ ( Anonymous314) https://math.stackexchange.com/questions/413683/show-that-frac-1-sqrtxy-frac-1-sqrtyz-frac-1-sqrtzx-geq2/417347

oranggggg wrote: parmenides51 wrote: Let $x,y,z\in \mathbb{R}^+_0$ such that $xy+yz+zx=1$. Prove that $$\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}\ge 2+\frac{1}{\sqrt{2}}.$$ ( Anonymous314) https://math.stackexchange.com/questions/413683/show-that-frac-1-sqrtxy-frac-1-sqrtyz-frac-1-sqrtzx-geq2/417347 math110@MSE's proof (I rewrote it compactly): WLOG, assume that $x \ge y \ge z$. Using $(x+y)(y+z) = xy + yz + zx + y^2 = 1 + y^2$ and $(y+z)(z+x) = xy + yz + zx + z^2 = 1 + z^2$, we have \begin{align*} &\frac{1}{\sqrt{x + y}} + \frac{1}{\sqrt{y + z}} + \frac{1}{\sqrt{z + x}}\\ =\,& \frac{1}{\sqrt{y + z}} + \sqrt{y+z}\left(\frac{1}{\sqrt{(x+y)(y+z)}} + \frac{1}{\sqrt{(y+z)(z+x)}}\right)\\ =\, & \frac{1}{\sqrt{y + z}} + \sqrt{y + z} \left(\frac{1}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + z^2}}\right)\\ =\, & \frac{1}{\sqrt{y + z}} + \sqrt{y + z} \sqrt{\frac{1}{1 + y^2} + \frac{1}{1 + z^2} + \frac{2}{\sqrt{(1 + y^2)(1 + z^2)}}}\\ \ge\, & \frac{1}{\sqrt{y + z}} + \sqrt{y + z} \sqrt{1 + \frac{1}{1 + (y + z)^2} + \frac{2}{\sqrt{1 + (y + z)^2}}} \qquad (1)\\ =\, & \frac{1}{\sqrt{y + z}} + \sqrt{y + z}\left(1 + \frac{1}{\sqrt{1 + (y + z)^2}}\right) \qquad (2) \end{align*}where in (1) we have used (easy to prove using $y + z \le 2$ and $yz \le \frac13$) $$\frac{1}{y^2 + 1} + \frac{1}{z^2 + 1} - 1 - \frac{1}{1 + (y + z)^2} = \frac{yz[2 - 2yz - yz(y + z)^2]}{(y^2 + 1)(z^2 + 1)[1 + (y + z)^2]}\ge 0$$and $$1 + (y + z)^2 - (y^2 + 1)(z^2 + 1) = yz(2 - yz) \ge 0.$$ Using (2), it suffices to prove that $$\frac{1}{\sqrt{y + z}} + \sqrt{y + z}\left(1 + \frac{1}{\sqrt{1 + (y + z)^2}}\right) \ge 2 + \frac{1}{\sqrt{2}}$$or $$\frac{1}{\sqrt{y + z}} + \sqrt{y + z} + \frac{1}{\sqrt{\frac{1}{y+z} + y + z}} \ge 2 + \frac{1}{\sqrt{2}}. \qquad (3)$$ Letting $t = \frac{1}{\sqrt{y + z}} + \sqrt{y + z} \ge 2$, (3) is written as $$t + \frac{1}{\sqrt{t^2 - 2}} \ge 2 + \frac{1}{\sqrt 2}$$which is true. We are done.

parmenides51 wrote: Let $x,y,z\in \mathbb{R}^+_0$ such that $xy+yz+zx=1$. Prove that $$\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}\ge 2+\frac{1}{\sqrt{2}}.$$ ( Anonymous314) My proof is not nice. WLOG, assume that $x \ge y \ge z$. By AM-GM, we have $$\frac{1}{\sqrt{x + y}} = \frac{2\sqrt 2}{2\sqrt{(x + y)\cdot 2}} \ge \frac{2\sqrt 2}{x + y + 2}. \qquad (1)$$ Also, we have \begin{align*} &\frac{1}{\sqrt{y+z}} + \frac{1}{\sqrt{z + x}}\\ =\,&\sqrt{\frac{1}{y+z} + \frac{1}{z+x} + \frac{2}{\sqrt{(y+z)(z+x)}}}\\ =\,& \sqrt{\frac{x + y + 2z}{1 + z^2} + \frac{2}{\sqrt{1+z^2}}} \qquad (2) \end{align*}where we have used $(y+z)(z+x) = xy + yz + zx + z^2 = 1 +z^2$. From (1) and (2), it suffices to prove that $$\frac{2\sqrt 2}{x + y + 2} + \sqrt{\frac{x + y + 2z}{1 + z^2} + \frac{2}{\sqrt{1+z^2}}} \ge 2 + \frac{1}{\sqrt 2}$$or $$\frac{x + y}{1 + z^2} + \frac{2z}{1 + z^2} + \frac{2}{\sqrt{1+z^2}} \ge \left(2 + \frac{1}{\sqrt 2} - \frac{2\sqrt 2}{x + y + 2}\right)^2.\qquad (3)$$ Fact 1: It holds that $$\frac{x + y}{1 + z^2} + \frac{2}{\sqrt{1+z^2}} \ge x + y + \frac{2}{1 + 7z/10}.$$(The proof is given at the end.) By Fact 1 and (3), it suffices to prove that $$x + y + \frac{2z}{1+z^2} + \frac{2}{1 + 7z/10} \ge \left(2 + \frac{1}{\sqrt 2} - \frac{2\sqrt 2}{x + y + 2}\right)^2. \qquad (4)$$ (4) is true. Omitted here. Proof of Fact 1: From $x\ge y \ge z$ and $xy + yz + zx = 1$, we have $z \le \frac{\sqrt 3}{3}$. We have $\sqrt{1+z^2} \le 1 + \frac{2z}{7}$. It suffices to prove that $$\frac{x + y}{1 + z^2} + \frac{2}{1 + 2z/7} \ge x + y + \frac{2}{1 + 7z/10}$$or $$\frac{2}{1 + 2z/7} - \frac{2}{1 + 7z/10} \ge \frac{z^2(x + y)}{1+z^2}.$$ Using $x\ge y\ge z$ and $xy + yz + zx = 1$, we have $z(x+y) = zx + yz \le 2/3$. It suffices to prove that $$\frac{2}{1 + 2z/7} - \frac{2}{1 + 7z/10} \ge \frac{2z/3}{1+z^2}$$which is true. We are done.

parmenides51 wrote: Let $x,y,z\in \mathbb{R}^+_0$ such that $xy+yz+zx=1$. Prove that $$\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}\ge 2+\frac{1}{\sqrt{2}}.$$ (Anonymous314) https://artofproblemsolving.com/community/c6h384068p2130988