2. Given $x>0, y>0, a=x+y, b=$ $\sqrt{x^{2}-x y+y^{2}}, c=\lambda \sqrt{x y}$. If $a, b, c$ can be the lengths of the three sides of a triangle, then the range of the positive real number $\lambda$ is $\qquad$
$2.1b$.
If $a, b, c$ can be the lengths of the three sides of a triangle, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y+\sqrt{x^{2}-x y+y^{2}}>\lambda \sqrt{x y}, \\
\lambda \sqrt{x y}+\sqrt{x^{2}-x y+y^{2}}>x+y
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
\lambda\frac{x+y-\sqrt{x^{2}-x y+y^{2}}}{\sqrt{x y}} .
\end{array}\right.
\end{array}
$$

Notice that,
$$
x+y \geqslant 2 \sqrt{x y}, \sqrt{x^{2}-x y+y^{2}} \geqslant \sqrt{x y} .
$$

Therefore, $\lambda<3$.
$$
\begin{array}{l}
\text { Also } \frac{x+y-\sqrt{x^{2}-x y+y^{2}}}{\sqrt{x y}} \\
=\frac{3 \sqrt{x y}}{x+y+\sqrt{x^{2}-x y+y^{2}}} \\
\leqslant \frac{3 \sqrt{x y}}{3 \sqrt{x y}}=1 .
\end{array}
$$

Hence $1<\lambda<3$.