1

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.

Martin Sleziak
  • 53,687
Oussama Sarih
  • 955
  • 2
    It is not defined for $a=b=c=0$ so the domain isn’t closed; there is no reason to expect a minimum to exist. The infimum exists, which you show to be $\sqrt2$ correctly. – Macavity Oct 31 '17 at 14:20
  • A problem some people may be interested in: Let $a, b, c \ge 0$ with $(a+b)(b+c)(c+a)\ne 0$. Find the minimum of $f(a,b,c) = \sqrt{\frac{a}{b+c}} + \sqrt[3]{\frac{b}{c+a}} + \sqrt[4]{\frac{c}{a+b}}$. (the minimum of $f$ is $\frac{3}{\sqrt[3]{4}}$) See: https://math.stackexchange.com/questions/1589425/finding-the-minimum-value-of-a-radical-expression – River Li Nov 27 '20 at 10:11

1 Answers1

3

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!

Michael Rozenberg
  • 194,933