<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] 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.
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