Solve using Cauchy Schwarz Inequality Use Cauchy–Schwarz inequality to prove that for three positive reals $a, b, c$ such that $a + b + c \leqslant 3$ then $\frac{1}{\sqrt{a}} +\frac{1}{\sqrt{b}} +\frac{1}{\sqrt{c}} \geqslant 3$. (Use C-S for $\sqrt{a} , \sqrt{b} , \sqrt{c}$ and $\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c} }$).
How do I solve this?

From Cauchy Schwarz inequality we have that $\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}$ and $\sqrt{a},\sqrt{b},\sqrt{c}$
Then $$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq (\frac{1}{\sqrt{a}} \cdot \sqrt{a}+\frac{1}{\sqrt{b}} \cdot \sqrt{b}+\frac{1}{\sqrt{c}} \cdot \sqrt{c})^2=(1+1+1)^2=9$$ 
So
$$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$$
In another hand for any $x,y,z$ we have $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$
because after expending you get that
$$x^2+y^2+z^2\geq xy+yz+zx$$
So back to our question we have that for $x=\sqrt{a},y=\sqrt{b},z=\sqrt{c}$
$$3(a+b+c)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ since a+b+c=3 and getting ride of square  then 
$$3\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})$$ 
From first inequality we have that 
$$3 (\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})\geq (\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$$
or 
$$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq 3$$
