Let $s=a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}$, $t=a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=b_{1}b_{2}+b_{2}b_{3}+b_{3}b_{4}$ and $p_{1}=a_{1}a_{2}a_{3}$, $p_{2}=b_{1}b_{2}b_{3}$. We also let \begin{align*} f\left(x\right) & =\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\\ & =x^{3}-sx^{2}+tx-p_{1}\\ g\left(x\right) & =\left(x-b_{1}\right)\left(x-b_{2}\right)\left(x-b_{3}\right)\\ & =x^{3}-sx^{2}+tx-p_{2} \end{align*}Then, polynomials $g\left(x\right)$ is negative on $[-\infty,b_{1}]$ and $[b_{2},b_{3}]$ (1). Therefore, we have $g\left(a_{1}\right)\leq0$. Then, we have $0=f\left(a_{1}\right)=g\left(a_{1}\right)+p_{2}-p_{1}$ or $p_{1}=f\left(a_{1}\right)+p_{2}\leq p_{2}$. Now, we assume $a_{3}>b_{3}$. Then, we have $g\left(a_{3}\right)>0$ from (1), implies $0=f\left(a_{3}\right)=g\left(a_{3}\right)+p_{2}-p_{1}$ or $p_{1}=g\left(a_{3}\right)+p_{2}>p_{2}$, a contradiction. Therefore, $a_{3}\leq b_{3}$.