$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
 A: Using Hölder's inequality we have:
$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2/3} (a(a+b)+b(b+c)+c(c+a))^{1/3}\geq a+b+c.$$
i.e.
$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2} \geq \frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca}.$$ 
We have to prove that: 
$$\frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{9}{2}.$$
i.e.
$$2(a+b+c)^3\geq9\left(a^2+b^2+c^2+ab+bc+ca\right). \tag{1}$$
Let $p=a+b+c$ and $q=ab+bc+ca$ and using that $abc=1$ and $AM-GM$ we obtain that $q \geq 3$. 
Inequality $(1)$ is equivalent to: 
$$2p^3 \geq 9\left(p^2-2q+q\right) \Leftrightarrow 2p^3+9q \geq 9p^2.$$
Applying $AM-GM$ we obtain 
$$2p^3+9q \geq 2p^3+27=p^3+p^3+27 \geq 3\cdot \sqrt[3]{27p^6}=9p^2,$$ as required.
A: Let $a=b$. Then $c=\frac{1}{a^2}$ and the formula is :
$$f(a)=\sqrt{\frac{a}{2}}+\sqrt{a+\frac{1}{a^2}}$$
$$f'(a)=\frac{1}{2\sqrt{2a}}+(\frac{1}{2}-\frac{1}{a^3}).\frac{1}{\sqrt{a+\frac{1}{a^2}}}$$
The only root in $[0,\infty)$ of f'(a) is 1, hence $f(1)=\frac{3}{\sqrt{2}}$ is a minimum.
Now, what happens if $a\neq b$ ?
Use the same method, but say $b=k.a+(1-k)$ (so let $k=\frac{b-1}{a-1}$ be a constant, and if $a=1$ exchange $a$ and $c$). Then $c=\frac{1}{ka^2+a(1-k)}$ and :
$$f_k(a)=\frac{a}{\sqrt{a.(1+k)+(1-k)}}+\frac{ka+(1-k)}{\sqrt{ka+(1-k)+\frac{1}{ka^2+a(1-k)}}}+\frac{\frac{1}{ka^2+a(1-k)}}{\sqrt{a+\frac{1}{ka^2+a(1-k)}}}$$
Once again, obtain $f'_k$ and show that its only root is $1$ (This is quite technical, use mathematica ?). I agree there should have something more simple.
A: 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.
A: Another way.
Let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}=\sum_{cyc}\frac{\frac{y}{x}}{\sqrt{\frac{y}{x}+\frac{z}{y}}}=\sum_{cyc}\frac{\sqrt{y^3}}{\sqrt{x(y^2+xz)}}=\sum_{cyc}\frac{y^2}{\sqrt{xy(y^2+xz)}}\geq$$
$$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}\sqrt{xy(y^2+xz)}}=\frac{2\sqrt2(x+y+z)^2}{2\sum\limits_{cyc}\sqrt{2xy\cdot(y^2+xz)}}\geq\frac{2\sqrt2(x+y+z)^2}{\sum\limits_{cyc}(2xy+y^2+xz)}=$$
$$=\frac{2\sqrt2(x+y+z)^2}{\sum\limits_{cyc}(x^2+3xy)}\geq\frac{2\sqrt2(x+y+z)^2}{\sum\limits_{cyc}\left(x^2+\frac{1}{3}x^2+\frac{8}{3}xy\right)}=\frac{2\sqrt2(x+y+z)^2}{\frac{4}{3}(x+y+z)^2}=\frac{3}{\sqrt2}.$$
