35. The left side of the inequality is easy to prove:
$$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$$

Now we prove the right side.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} \left(\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}\right)^{2}= \\ {\left[\sqrt{a^{2}+c^{2}} \cdot \frac{a}{\sqrt{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}}+\sqrt{b^{2}+a^{2}} \cdot \frac{b}{\sqrt{\left(b^{2}+c^{2}\right)\left(b^{2}+a^{2}\right)}}+\right.} \\ \left.\sqrt{c^{2}+b^{2}} \cdot \frac{c}{\sqrt{\left(c^{2}+a^{2}\right)\left(c^{2}+b^{2}\right)}}\right]^{2} \leqslant \\ {\left[\left(a^{2}+c^{2}\right)+\left(b^{2}+a^{2}\right)+\left(c^{2}+b^{2}\right)\right] \cdot} \\ {\left[\frac{a^{2}}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}+\frac{b^{2}}{\left(b^{2}+c^{2}\right)\left(b^{2}+a^{2}\right)}+\frac{c^{2}}{\left(c^{2}+a^{2}\right)\left(c^{2}+b^{2}\right)}\right]=} \\ 2\left(a^{2}+b^{2}+c^{2}\right)\left[\frac{a^{2}}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}+\right. \\ \left.\frac{b^{2}}{\left(b^{2}+c^{2}\right)\left(b^{2}+a^{2}\right)}+\frac{c^{2}}{\left(c^{2}+a^{2}\right)\left(c^{2}+b^{2}\right)}\right] \end{array}$$

Now we prove
$$\begin{array}{l} \frac{a^{2}\left(a^{2}+b^{2}+c^{2}\right)}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}+\frac{b^{2}\left(a^{2}+b^{2}+c^{2}\right)}{\left(b^{2}+c^{2}\right)\left(b^{2}+a^{2}\right)}+\frac{c^{2}\left(a^{2}+b^{2}+c^{2}\right)}{\left(c^{2}+a^{2}\right)\left(c^{2}+b^{2}\right)} \leqslant \frac{9}{8} \\ (1) \Leftrightarrow 8\left(a^{2}+b^{2}+c^{2}\right)\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \leqslant 9\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+\right. \end{array}$$
$\left.a^{2}\right) \Leftrightarrow a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}+a^{2} b^{4}+b^{2} c^{4}+c^{2} a^{4} \geqslant 6 a^{2} b^{2} c^{2}$. This follows from the AM-GM inequality.
Thus, $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2}$.