<s>[INST] Find the least value of $\frac{1}{x}+\frac{3}{y}+\frac{5}{z}$ Given that $x$,$y$ and $z$ are positive numbers and $x+y+z=1$ , Find the least value of $\frac{1}{x}+\frac{3}{y}+\frac{5}{z}$.
My approach:
We write,
$$\frac{1}{x}+\frac{3}{y}+\frac{5}{z}=\frac{x+y+z}{x}+\frac{(3)(x+y+z)}{y}+\frac{(5)(x+y+z)}{z}$$
This further simplifies to,
$$ 9+\left(\frac{y}{x}+\frac{3x}{y}\right)+\left(\frac{z}{x}+\frac{5x}{z}\right)+\left(\frac{3z}{y}+\frac{5y}{z}\right)$$
Now we can use AM-GM inequalitie in each of the parentheses.
So the least value of above expression works out to
$$9+2\left(\sqrt{3}+\sqrt{5}+\sqrt{15}\right)$$
Is this correct?? If not could you please suggest some ways of approaching this problem?.
Thank you!
 [/INST] One can show that $\sum \limits_{I=1}^{n}\dfrac{a_i^2}{b_i}\geq \dfrac{(a_1+\dots+a_n)^2}{b_1+\dots+b_n}.$ using this inequality it follows that $$\frac{1}{x}+\frac{3}{y}+\frac{5}{z}\geq (1+\sqrt{3}+\sqrt{5})^2$$
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