3. Given real numbers $a, b, c$ and $\lambda>0$ satisfy
$$
\begin{array}{l}
|b+\lambda c| \geqslant(\lambda+1)|a|, \\
|c+\lambda a| \geqslant(\lambda+1)|b|, \\
|a+\lambda b| \geqslant(\lambda+1)|c| .
\end{array}
$$

Then the relationship between $a, b, c$ is $\qquad$
3. $a=b=c$.

Squaring both sides of the three inequalities, we have
$$
\begin{array}{l}
b^{2}+2 \lambda b c+\lambda^{2} c^{2} \geqslant\left(\lambda^{2}+2 \lambda+1\right) a^{2}, \\
c^{2}+2 \lambda c a+\lambda^{2} a^{2} \geqslant\left(\lambda^{2}+2 \lambda+1\right) b^{2}, \\
a^{2}+2 \lambda a b+\lambda^{2} b^{2} \geqslant\left(\lambda^{2}+2 \lambda+1\right) c^{2} .
\end{array}
$$

Adding and simplifying the above three equations, we get
$$
2 \lambda(a b+b c+c a) \geqslant 2 \lambda\left(a^{2}+b^{2}+c^{2}\right) \text {. }
$$

Since $\lambda>0$, then $2(a b+b c+c a) \geqslant 2\left(a^{2}+b^{2}+c^{2}\right)$, which means $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \leqslant 0$.
Therefore, $a=b=c$.