For $a,b,c>0$. Minimize $P=a+b+c$

Question

For $a,b,c>0$ and $\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1$, minimize $$P=a+b+c$$

Answer 1

By Cauchy Schwarz: $$ 1\times P=\left(\frac{2}{a}+\frac{5}{b}+\frac{3}{c}\right)(a+b+c)\geq(\sqrt{2}+\sqrt{5}+\sqrt{3})^2 $$ so $\min P=(\sqrt{2}+\sqrt{5}+\sqrt{3})^2$ when $$ \frac{2}{a^2}=\frac{5}{b^2}=\frac{3}{c^2}\quad\text{and}\quad\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1; $$ i.e. $$ a=2+\sqrt{6}+\sqrt{10},\quad b=5+\sqrt{10}+\sqrt{15},\quad c=3+\sqrt{6}+\sqrt{15}. $$

Answer 2

Let $x=\frac2a,y=\frac5b,z=\frac3c.$ $$\min P=\min\left(\left\{\left.\frac2x+\frac5y+\frac3{1-x-y}\,\right|\,x,y>0,x+y<1\right\}\right)$$ is attained at the critical point, $$x=\frac{\sqrt2}{\sqrt2+\sqrt3+\sqrt5},\quad y=\frac{\sqrt5}{\sqrt2+\sqrt3+\sqrt5},$$ hence its value is $$2\frac{\sqrt2+\sqrt3+\sqrt5}{\sqrt2}+5\frac{\sqrt2+\sqrt3+\sqrt5}{\sqrt5}+3\frac{\sqrt2+\sqrt3+\sqrt5}{\sqrt3}$$ $$=(\sqrt2+\sqrt3+\sqrt5)^2.$$