<s>[INST] Minimum of an expression involving radicals: $\sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}}$ The link below is a MSE question discussing the maximum value of the expression 
$$ \sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}} $$ 
which is $3$.
Prove inequality $\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3$
I thought about the minimum of the same expression over nonnegative reals.
My reasoning was as follows :
Using the fact that $\sqrt{X+Y+Z} \le \sqrt X + \sqrt Y + \sqrt Z$, and the fact $u + v \le u+v+w$ we can already assert that the expression to minimize is $\ge  \sqrt 2$
Now taking $a = 0$, $b\to 0^+$ and $c$ arbitrary we see that the expression can take values arbitrarily close to $\sqrt 2$. Thus $\sqrt 2 $ is the infimum. but It seems not to be a minimum while the continuous function we have must have a minimum on the compact sphere (homogenous expression). Is there anything wrong with this approach ?
Then, What would the result be if we restrict ourselves to positive reals.
Thanks.
 [/INST] For $b=\epsilon^2$ and $c=\epsilon$, where $\epsilon\rightarrow0^+$ our expression is closed to $\sqrt2$.
We'll prove that for positive variables it's an infimum. 
Indeed, we need to prove that
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}\geq1,$$ which is true by Holder:
$$\left(\sum_{cyc}\sqrt{\frac{a}{a+b}}\right)^2\sum_{cyc}a^2(a+b)\geq(a+b+c)^3.$$
Thus, it remains to prove that
$$(a+b+c)^3\geq\sum_{cyc}a^2(a+b)$$ or
$$\sum_{cyc}(2a^2b+3a^2c+2abc)\geq0,$$ which is obvious.
Done!
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