已知正数 $a,b$ 满足 $a+b=1$，求 $M=\sqrt{1+2a^2}+2\sqrt{\left(\dfrac 5{12}\right)^2+b^2}$ 的最小值．
答案 $\dfrac{5\sqrt{34}}{12}$．
解析 引入参数 $\lambda,\mu$（$\lambda,\mu>0$），根据柯西不等式，有\[\begin{cases} \sqrt{1+2a^2}\geqslant \dfrac{\lambda+a}{\sqrt{\lambda^2+\dfrac 12}}\\ 2\sqrt{\left(\dfrac5{12}\right)^2+b^2}\geqslant \dfrac{\dfrac {5\mu}6+2b}{\sqrt{\mu^2+1}},\end{cases}\]于是\[M\geqslant \dfrac{\lambda}{\sqrt{\lambda^2+\dfrac 12}}+\dfrac{5\mu}{6\sqrt{\mu^2+1}}+\dfrac{1}{\sqrt{\lambda^2+\dfrac 12}}\cdot a+\dfrac{2}{\sqrt{\mu^2+1}}\cdot b,\]考虑取等条件可得\[\begin{cases} 2a=\dfrac{1}{\lambda},\\ b=\dfrac{5}{12\mu},\\ a+b=1,\\ \dfrac{1}{\sqrt{\lambda^2+\dfrac 12}}=\dfrac2{\sqrt{\mu^2+1}},\end{cases}\iff \begin{cases} \dfrac{1}{2\lambda}+\dfrac{5}{12\mu}=1,\\ \mu^2=4\lambda^2+1,\\ (a,b)=\left(\dfrac{1}{2\lambda},\dfrac{5}{12\mu}\right),\end{cases}\iff \begin{cases} \left(\lambda,\mu\right)=\left(\dfrac 23,\dfrac 53\right),\\ \left(a,b\right)=\left(\dfrac 34,\dfrac 14\right),\end{cases}\]从而所求代数式的最小值当 $\left(a,b\right)=\left(\dfrac 34,\dfrac 14\right)$ 时取得，为 $\dfrac{5\sqrt{34}}{12}$．