1738

已知正实数 ${\displaystyle a,b}$ 满足 ${\displaystyle \frac{2}{a}+\frac{4}{b}=1}$, 求 ${\displaystyle a+b+\sqrt{a^2+b^2}}$ 的最小值.

注意到^[1] $$(a^2+b^2)(9+16)\ge (3a+4b{)}^2,$$ 于是 $$\begin{array}{rl}\text{原式}& \ge a+b+\frac{3a+4b}{5}=\frac{1}{5}(8a+9b)\\ & =\frac{1}{5}(8a+9b)(\frac{2}{a}+\frac{4}{b})=\frac{1}{5}(52+\frac{32a}{b}+\frac{18b}{a})\ge 20,\end{array}$$ 取等条件为 $$\{\begin{array}{rl}& \frac{a}{b}=\frac{3}{4},\\ & \frac{2}{a}+\frac{4}{b}=1\\ & \frac{32a}{b}=\frac{18b}{a}\end{array}\Rightarrow (a,b)=(5,\frac{20}{3}).$$

设 ${\displaystyle a=r\cos \theta }$, ${\displaystyle b=r\sin \theta }$, ${\displaystyle \theta \in (0,\frac{\pi }{2})}$, 则根据条件式 ${\displaystyle r=\frac{2}{\cos \theta }+\frac{4}{\sin \theta }}$. 现在记 ${\displaystyle t=\tan \frac{\theta }{2}}$, 则根据万能公式 $$\begin{array}{rl}\text{原式}& =r(\cos \theta +\sin \theta +1)=(\frac{2(1+t^2)}{1-t^2}+\frac{4(1+t^2)}{2t})(\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}+1)\\ & =4(\frac{1}{1-t}+\frac{1}{t}+1)\ge 4(\frac{(1+1{)}^2}{(1-t)+t}+1)=20.\end{array}$$ 等号成立当且仅当 ${\displaystyle t=\frac{1}{2}}$.