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.
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!