Irska

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 $$a\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{a}(b+c)<3\sqrt{2}⟺\left(\frac{a}{bc}+\frac{1}{a}\right)(b+c)<3\sqrt{2}.$$ 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 $$\begin{array}{rl}\left(\frac{a}{bc}+\frac{1}{a}\right)(b+c)& \ge \left(\frac{\sqrt{2}}{\sqrt{bc}}+\frac{1}{\sqrt{2\sqrt{bc}}}\right)(b+c)\\ & =\frac{3}{\sqrt{2\sqrt{bc}}}(b+c)\\ & \ge \frac{6\sqrt{bc}}{\sqrt{2}\sqrt{bc}}\\ & =3\sqrt{2},\end{array}$$ 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.