<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] 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.
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