Finding the minimum value of $\sqrt { \frac { a }{ b+c } } +\sqrt [ 3 ]{ \frac { b }{ c+a } } +\sqrt [ 4 ]{ \frac { c }{ a+b } }$ If $a, b, c\ge 0$ with $(a+b)(b+c)(c+a) > 0$,
find the minimum of $\sqrt { \frac { a }{ b+c }  } +\sqrt [ 3 ]{ \frac { b }{ c+a }  } +\sqrt [ 4 ]{ \frac { c }{ a+b }  }$.
The minimum is $\frac{3}{\sqrt[3]{4}}$ achieved at $b = 0, \frac{a}{c} = 2^{-4/3}$.
I am not able to progress in this problem.I tried applying AM-GM,Cauchy,Weighted AM-GM,etc. but none seem to provide fruitful results. Please help.
Source: A collection of problems which couldn't be solved by any teacher of my school.
Thanks.
 A: Remark: Here is an ugly solution. Hope to see nice solutions.
Problem: Let $a, b, c \ge 0$ with $(a+b)(b+c)(c+a)\ne 0$. Find the minimum of
$f(a,b,c) = \sqrt{\frac{a}{b+c}} + \sqrt[3]{\frac{b}{c+a}} + \sqrt[4]{\frac{c}{a+b}}$.
Solution:
If $b=0$, by AM-GM inequality, we have $f = \sqrt{\frac{a}{c}} + \sqrt[4]{\frac{c}{a}} \ge \frac{3}{\sqrt[3]{4}}$
with equality if $\frac{a}{c} = 2^{-4/3}$.
Let us prove that the minimum of $f$ is $\frac{3}{\sqrt[3]{4}}$. It suffices to prove that, for $a, c \ge 0$ and $a+c > 0$,
$$\sqrt{\frac{a}{1+c}} + \sqrt[3]{\frac{1}{c+a}} + \sqrt[4]{\frac{c}{a+1}} \ge \frac{3}{\sqrt[3]{4}}$$
or
$$\sqrt{\frac{2a\sqrt[3]{2}}{1+c}} + \sqrt[3]{\frac{4}{c+a}} + \sqrt[4]{\frac{8c}{(a+1)\sqrt[3]{2}}} \ge 3.$$
Let
$$\frac{2a\sqrt[3]{2}}{1+c} = y^2, \quad \frac{8c}{(a+1)\sqrt[3]{2}} = x^4.$$
Correspondingly, we have
$$a = \frac{y^2(x^4\sqrt[3]{2} + 8)}{(16-x^4y^2)\sqrt[3]{2}}, \quad
c = \frac{x^4(y^2 + 2\sqrt[3]{2})}{16-x^4y^2}.$$
The constraint is $x, y \ge 0; \ x^4y^2 < 16$. It suffices to prove that
$$y + \sqrt[3]{\frac{2\sqrt[3]{2}(16 - x^4y^2)}{x^4y^2\sqrt[3]{2} + x^4\sqrt[3]{4} + 4y^2}} + x \ge 3.$$
It suffices to prove that, for $x, y \ge 0; \ x^4y^2 < 16; \ x + y < 3$,
$$\frac{2\sqrt[3]{2}(16 - x^4y^2)}{x^4y^2\sqrt[3]{2} + x^4\sqrt[3]{4} + 4y^2} \ge (3-x-y)^3$$
which is written as (after clearing the denominators)
$$-x^4(3 - x - y)^3 \sqrt[3]{4} + g(x, y)\sqrt[3]{2} - 4y^2(3 - x - y)^3 \ge 0$$
where $g(x, y) = 32 - 2x^4y^2 - (3-x-y)^3x^4y^2$.
Consider
$$F(q) = -x^4(3 - x - y)^3 q^2 + g(x, y)q - 4y^2(3 - x - y)^3.$$
Since $F(q)$ is concave and $\frac{5}{4} < \sqrt[3]{2} < \frac{19}{15}$, to prove $F(\sqrt[3]{2}) \ge 0$, it suffices to prove that
$$F(\tfrac{5}{4}) \ge 0, \quad F(\tfrac{19}{15}) \ge 0.$$
It suffices to prove that, for $x, y\ge 0$ and $x+y \le 3$,
$$F(\tfrac{5}{4}) \ge 0, \quad F(\tfrac{19}{15}) \ge 0.$$
They are verified by Mathematica. There are ugly proofs. Omitted.
