<s>[INST] $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
 [/INST] The function $$f(x)=\frac{1}{\sqrt{x}}$$ is convex. Applying Jensen as follows :
$$
a*f[a + b] + b*f[b + c] + c*f[c + a] >= (a + b + c)*f[\frac{
  (a (a + b) + b (b + c) + c (c + a))}{(a + b + c)}] = \frac{(a + b + c)^{3/2}}{\sqrt{a^2 + a b + b^2 + a c + b c + c^2}}
$$
We need to prove $$\frac{(a + b + c)^{3/2}}{\sqrt{a^2 + a b + b^2 + a c + b c + c^2}}>=\frac{3}{\sqrt{2}}$$
This is equivalent to proving:
$$2 (a + b + c)^3 - 9 (a^2 + a b + b^2 + a c + b c + c^2)>=0$$
Which is an easy exercise.
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