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?
 A: 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 
$$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq 3$$
A: This looks more like the identity between the harmonic and quadratic mean
$$
\left(\frac{x_1^{-1}+x_2^{-1}+…+x_n^{-1}}n\right)^{-1}\le\left(\frac{x_1^2+x_2^2+…+x_n^2}n\right)^{\tfrac12}
$$
with $n=3$, $x_1=\sqrt a$, $x_2=\sqrt b$, $x_3=\sqrt c$.

Or you can apply CSI twice
$$
n=x_1\frac1{x_1}+x_2\frac1{x_2}+…+x_n\frac1{x_n}\le\sqrt{x_1^2+x_2^2+…+x_n^2}\sqrt{\frac1{x_1^2}+\frac1{x_2^2}+…+\frac1{x_n^2}}
\\
\le\sqrt{\sqrt{n}\sqrt{x_1^4+x_2^4+…+x_n^4}}\sqrt{\frac1{x_1^2}+\frac1{x_2^2}+…+\frac1{x_n^2}}
$$
with $x_1=\sqrt[4]a$ etc. This, after squaring and replacing $y_k=x_k^2$ gives again the harmonic-quadratic mean inequality.
