Recently I answered this Question which required a proof for the upper and lower bound of a given function. While seemingly a calculus question, I realised that it can be transformed into an inequality problem, and my solution is detailed there. However, it seems that we can improve upon the upper bound substantially, from $2$ to $\sqrt{3}$. This leads us to the following claim:
Let $x,y,z \in \mathbb{R^+},xyz=1$. Prove that $\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq \sqrt{3}$.
Equivalently, by the substitution $x=\dfrac{b}{a}, y=\dfrac{c}{b}, z=\dfrac{a}{c}$, it suffices for us to prove that for all $a,b,c \in \mathbb{R^+}$, $\sqrt{\dfrac{a}{a+2b}} + \sqrt{\dfrac{b}{b+2c}} + \sqrt{\dfrac{c}{c+2a}} \leq \sqrt{3}. $ But after thinking for a while, I still have no clue on how to proceed.
Any help will be much appreciated.
Your inequality doesn't true because the maximal value of k for which the inequality $$\dfrac{1}{\sqrt{1+kx}} + \dfrac{1}{\sqrt{1+ky}} + \dfrac{1}{\sqrt{1+kz}} \leqslant \frac{3}{\sqrt{k+1}},$$ is true for all positive real numbers $x,\,y,\,z$ satisfy $xyz=1$ is $k = \frac54.$