Proof: Since $x_1$, $x_2$, $x_3$ are positive real numbers,

we have $$\frac {x_{2}^{2}}{x_{1}}+x_{1}\geq2x_{2}$$, $$\frac {x_{3}^{2}}{x_{2}}+x_{2}\geq2x_{3}$$, $$\frac {x_{1}^{2}}{x_{3}}+x_{3}\geq2x_{1}$$,

Adding the three inequalities, we get $$\frac {x_{2}^{2}}{x_{1}}+x_{1}+ \frac {x_{3}^{2}}{x_{2}}+x_{2}+ \frac {x_{1}^{2}}{x_{3}}+x_{3}\geq2(x_{1}+x_{2}+x_{3})$$,

Since $x_1+x_2+x_3=1$, we have $$\frac { x_{ 2 }^{ 2 }}{x_{1}}+ \frac { x_{ 3 }^{ 2 }}{x_{2}}+ \frac { x_{ 1 }^{ 2 }}{x_{3}}\geq1$$.

Thus, the proof is complete, and the final answer is $\boxed{\geq1}$.