Let $ AC_1 = AB_1 = x, BC_1 = BA_1 = y$ ,and $ CA_1 = CB_1 = z$ It's equivalent to prove that $ \sum_{cyc}\sqrt {\frac {x}{x + y}} \le \frac {3}{\sqrt {2}}$ which is well-known, \[ \sum_{cyc}\sqrt {\frac {x}{x + y}} = \sum\frac {\sqrt {(xy + xz)(z + x)}}{\sqrt {(x + y)(y + z)(z + x)}} \le \frac {\sqrt {2(xy + yz + zx)}\sqrt {2(x + y + z)}}{\sqrt {(x + y)(y + z)(z + x)}} = 2\sqrt {1 + \frac {xyz}{(x + y)(y + z)(z + x)}} \le \frac {3}{\sqrt {2}} \]

rewrite it as $ \sum \sqrt{\dfrac{1}{1+m^2}}\leq \dfrac{3}{\sqrt 2}$, where $ m,n,p$ are positive reals whose product is $ 1$... suppose wlog that $ mn\leq 1$... it's easy to prove that $ \sqrt{\dfrac{1}{1+m^2}}+\sqrt{\dfrac{1}{1+n^2}}\leq 2\sqrt{\dfrac{1}{1+mn}}$ if $ mn\leq 1$... then, we're left to prove that $ 2\sqrt{\dfrac{1}{1+t}}+\sqrt{\dfrac{t^2}{1+t^2}}\leq \dfrac{3}{\sqrt 2}$... after squaring it once, it follows it's equivalent to $ 7t^3-t^2+9t+1\geq 8t\sqrt{(t+1)(t^2+1)}$... from am-gm we have that $ 4t(t^2+t+2)\geq 8t\sqrt{(t+1)(t^2+1)}$ and $ 7t^3-t^2+9t+1\geq 4t(t^2+t+2)$, so we're done

campos wrote: rewrite it as $ \sum \sqrt {\dfrac{1}{1 + m^2}}\leq \dfrac{3}{\sqrt 2}$, where $ m,n,p$ are positive reals whose product is $ 1$... suppose wlog that $ mn\leq 1$... it's easy to prove that $ \sqrt {\dfrac{1}{1 + m^2}} + \sqrt {\dfrac{1}{1 + n^2}}\leq 2\sqrt {\dfrac{1}{1 + mn}}$ if $ mn\leq 1$... then, we're left to prove that $ 2\sqrt {\dfrac{1}{1 + t}} + \sqrt {\dfrac{t^2}{1 + t^2}}\leq \dfrac{3}{\sqrt 2}$... after squaring it once, it follows it's equivalent to $ 7t^3 - t^2 + 9t + 1\geq 8t\sqrt {(t + 1)(t^2 + 1)}$... from am-gm we have that $ 4t(t^2 + t + 2)\geq 8t\sqrt {(t + 1)(t^2 + 1)}$ and $ 7t^3 - t^2 + 9t + 1\geq 4t(t^2 + t + 2)$, so we're done very nice inequality, put that $ a^2=\frac{x}{y}$ ... the inequality is equiavalent as: $ \sum \frac{1}{\sqrt{1+a^2}} \leq \frac{3}{\sqrt{2}}$ with $ abc=1$ the inequality is symetric then we can assume that $ a\geq b\geq c$ the $ bc\leq1$ From Cauchy Schwarz we have: $ \frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \sqrt{2}.\sqrt{\frac{1}{1+b^2}+\frac{1}{1+c^2}} =\sqrt{2}.\sqrt{1+\frac{1-b^2c^2}{b^2+1)(c^2+1}} \leq \sqrt{2}.\sqrt{1+\frac{1-b^2c^2}{(bc+1)^2}}= \frac{2}{\sqrt{1+bc}}$ and: $ \frac{1}{\sqrt{1+a^2}} \leq \frac{\sqrt{2}}{1+a}= \frac{bc\sqrt{2}}{bc+1}$ then it's suffice to prove that: $ \frac{bc\sqrt{2}}{bc+1} + \frac{2}{\sqrt{1+bc}} \leq \frac{3}{\sqrt{2}}$ $ \Leftrightarrow \left(\sqrt{\frac{2}{1+bc}}-1\right)^2 \geq 0$