Problem 1230

Prove the right inequality first: By the Pigeonhole Principle, $\frac{a}{\sqrt{a^{2}+b^{2}}}, \frac{b}{\sqrt{b^{2}+c^{2}}}, \frac{c}{\sqrt{c^{2}+a^{2}}}$, at least two of them are on the same side of $\frac{\sqrt{2}}{2}$ (or at $\frac{\sqrt{2}}{2}$). Without loss of generality, assume $\frac{a}{\sqrt{a^{2}+b^{2}}}, \frac{b}{\sqrt{b^{2}+c^{2}}}$ are on the same side of $\frac{\sqrt{2}}{2}$ (or at $\frac{\sqrt{2}}{2}$),
then $\left(\frac{a}{\sqrt{a^{2}+b^{2}}}-\frac{\sqrt{2}}{2}\right)\left(\frac{b}{\sqrt{b^{2}+c^{2}}}-\frac{\sqrt{2}}{2}\right) \geqslant 0$,
which means $\frac{\sqrt{2}}{2}\left(\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}\right) \leqslant \frac{a}{\sqrt{a^{2}+b^{2}}} \cdot \frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{1}{2}$.

Thus, $\frac{\sqrt{2}}{2}\left(\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}\right)$
$\leqslant \frac{1}{2}+\frac{a}{\sqrt{a^{2}+b^{2}}} \cdot \frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{2\left(c^{2}+a^{2}\right)}}$
Notice that $\sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)} \geqslant a b+b c, \sqrt{2\left(c^{2}+a^{2}\right)} \geqslant c+a$, then we have
$\frac{1}{2}+\frac{a}{\sqrt{a^{2}+b^{2}}} \cdot \frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{2\left(c^{2}+a^{2}\right)}} \leqslant \frac{1}{2}+\frac{a b}{a b+b c}+\frac{c}{c+a}=\frac{3}{2}$.
Therefore, $\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2}$.
Equality holds if and only if $a=b=c$.
Now prove the left inequality: By $a^{2}+b^{2}>a^{2}>0$, we have $0<\frac{a^{2}}{a^{2}+b^{2}}>\frac{a^{2}}{a^{2}+b^{2}+c^{2}}$,
which means $\frac{a}{\sqrt{a^{2}+b^{2}}}>\frac{a^{2}}{a^{2}+b^{2}+c^{2}}$.
Similarly, $\frac{b}{\sqrt{b^{2}+c^{2}}}>\frac{b^{2}}{a^{2}+b^{2}+c^{2}}, \frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{c^{2}}{a^{2}+b^{2}+c^{2}}$.
Thus, $\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{a^{2}+b^{2}+c^{2}}{a^{2}+b^{2}+c^{2}}=1$.
Hence, the original inequality is proved.