By using Cauchy-Schwartz inequality twice, we have \[\frac1{\sqrt{x^2+m^2+a^2}}\le\frac1{\frac1{\sqrt3}(x+m+a)}\le\frac{\sqrt3}9\left(\frac1x+\frac1m+\frac1a\right).\]Taking a cyclic sum, we get \begin{align*} \sum_{\rm cyc}\frac1{\sqrt{x^2+m^2+a^2}}&\le\sum_{\rm cyc}\frac{\sqrt3}9\left(\frac1x+\frac1m+\frac1a\right)=\frac{\sqrt3}3\sum_{\rm cyc}\frac1x\\ &=\frac{\sqrt3}3\cdot\frac{xma+mas+asx+sxm}{xmas}=\frac1{xmas}. \end{align*}

From QM-HM, we have \begin{align*} \sqrt{\dfrac{x^2+m^2+a^2}{3}} &\ge \dfrac{3}{\dfrac{1}{x}+\dfrac{1}{m}+\dfrac{1}{a}} \\ \implies \dfrac{1}{\sqrt{x^2+m^2+a^2}} &\le \dfrac{\dfrac{1}{x}+\dfrac{1}{m}+\dfrac{1}{a}}{3\sqrt{3}}. \\ \end{align*}Summing cyclically, we get \begin{align*} \sum_{cyc}\dfrac{1}{x^2+m^2+a^2} &\le \dfrac{\sum_{cyc}\dfrac{1}{x}}{\sqrt{3}} \\ &= \dfrac{\dfrac{\sum_{cyc}xma}{xmas}}{\sum_{cyc}xma} \\ &= \dfrac{1}{xmas}. \end{align*}We are done.