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

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    Lagrange's multipliers seem the way to go. – 2017-02-20
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    we should start with...? – 2017-02-20
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    ... Lagrange's multipliers. – 2017-02-20
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    can you use other way ? – 2017-02-20
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    Obviously $a>2,b>5,c>3$.So we found one The minimum value of P is greater than 10 – 2017-02-20

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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. $$ a=2+\sqrt{6}+\sqrt{10},\quad b=5+\sqrt{10}+\sqrt{15},\quad c=3+\sqrt{6}+\sqrt{15}. $$