Problem 1338

Suppose this is false. Then, without loss of generality, we can assume that $a^2 \ge b^2 + c^2$. Since
$$(a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = 3 + a \left( \frac{1}{b} + \frac{1}{c} \right) + \frac{1}{a} (b+c) + \frac{b}{c} + \frac{c}{b},$$
and
$$\frac{b}{c} + \frac{c}{b} \ge 2,$$
the hypothesis tells us that
$$a \left( \frac{1}{b} + \frac{1}{c} \right) + \frac{1}{a} (b+c) < 3\sqrt{2} \iff \left( \frac{a}{bc} + \frac{1}{a} \right) (b+c) < 3\sqrt{2}.$$
But the function
$$x \mapsto \frac{x}{bc} + \frac{1}{x}$$
is strictly increasing on the interval $[\sqrt{bc}, \infty)$, and $a \ge \sqrt{b^2+c^2} \ge \sqrt{2bc}$.
Hence
$$\begin{aligned} \left(\frac{a}{bc} + \frac{1}{a}\right) (b+c) &\ge \left(\frac{\sqrt{2}}{\sqrt{bc}} + \frac{1}{\sqrt{2\sqrt{bc}}}\right) (b+c) \\ &= \frac{3}{\sqrt{2\sqrt{bc}}} (b+c) \\ &\ge \frac{6\sqrt{bc}}{\sqrt{2}\sqrt{bc}} \\ &= 3\sqrt{2}, \end{aligned}$$
which conflicts with the hypotheses. Thus $a^2 < b^2 + c^2$. Similarly, $b^2 < c^2 + a^2$, $c^2 < a^2 + b^2$. From these inequalities it follows easily that $a < b+c$, $b < c+a$, $c < a+b$, and from both sets of inequalities it follows that $a$, $b$, $c$ are the lengths of the sides of an acute-angled triangle.