Given triangle with vertices $A,B,C$ (sides $|AB|=c,|BC|=a,|AC|=b$) and also given that the angle at $A$, denoted by $\alpha$ is at most $\pi/3$. Moreover, we know that there is some positive $d$ such that the sides $b,c$ satisfy $b\leq d, c\leq d$. I want to show that the side $a$ must also satisfy $a\leq d$.
I tried approaching this by the law of cosines, writing $$a^2=b^2+c^2-2bc\cos \alpha \leq b^2+c^2-bc=(b-c)^2+bc \leq (b-c)^2+d^2$$ but then got stuck. It seems like i need to get rid of the term $(b-c)^2$ somehow. The first $\leq$comes from the fact that $\alpha \leq \frac{\pi}{3}$ thus $\cos \alpha \geq \frac{1}{2}$ i.e. $-\cos \alpha \leq -\frac{1}{2}$.
I am still thinking about your way. Meanwhile:
If both $\angle B$ and $\angle C$ are less than $60^\circ$ then $\angle A+\angle B+\angle C<180^\circ$. So, without loss of generality, $\angle B\geq 60^\circ$. This implies that $\angle B\geq\angle A.$ Hence, $a\leq b\leq d$ .
We can also look at the function $f(b,c)=b^2+c^2-bc$ and compute that the it is bounded by $d^2$ in the square $[0,d]^2$. See Wolfram Alpha.