The Answer is Not $\frac{ 3}{ \sqrt{2} }$

In this solution, we will find an expression that has the same range as the expression given in the question, and use this new form to clearly show the minimum value the expression can take.

Define the function $f$ with domain $\{(a,b,c)\mid a,b,c \gt 0, a,b,c \in \mathbb{R}\}$ as the left hand side of the given inequality. $f(a,b,c) = \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}}$ Define the function $g$ with domain $\{(x,y)\mid x,y\gt 0, x,y\in \mathbb{R}\}$ as follows. $g(x,y) = \sqrt{\frac{x}{xy+1}} + \sqrt{\frac{xy}{x+1}} + \sqrt{\frac{1}{x+xy}}$ For any point $(a,b,c)$ in the domain of $f$ , $f(a,b,c) = g\left(\frac{a}{c},\frac{b}{a}\right)$ For any point $(x,y)$ in the domain of $g$ , $g(x,y) = f(x,xy,1)$ Therefore, the range of $f$ is equal to the range of $g$ .

Now note that for any $x_0\lt x_1$ , we have $\frac{x_0}{x_0y+1}\gt\frac{x_1}{x_1y+1}$ , $\frac{x_0y}{x_0+1}\gt\frac{x_1y}{x_1+1}$ , and $\frac{1}{x_0+x_0y}\gt\frac{1}{x_1+x_1y}$ . Thus, $g(x_0,y)\gt g(x_1,y)$ Therefore, for any $x_0$ , $\begin{aligned} g(x_0,y) & \gt \lim_{x\to\infty}(g(x,y))\\ & = \sqrt{\frac{1}{y}} + \sqrt{\frac{y}{1}} + \sqrt{0} \end{aligned}$ Note that by the definition of a limit, we can get as close to $\sqrt{y} + \sqrt{\frac{1}{y}}$ as desired.

It remains to find the minimum of $h(y) = \sqrt{y} + \sqrt{\frac{1}{y}}$ with domain $y\gt0$ . $\begin{aligned} 0 & = h'(y) \\ & = \frac{y^{-1/2}}{2} - \frac{y^{-3/2}}{2} \\ y & = 1 \end{aligned}$ With a little bit of investigation, it is easy to see that the critical point at $y=1$ is a global minimum for the domain of $h$ .

Therefore, $f(a,b,c) \gt \sqrt{1} + \sqrt{1} = 2$ .

It is interesting to note that this is strictly greater than. The question asked for greater than or equal to, but as discussed, the nature of the limit still has 2 as the greatest lower bound, even though 2 is not actually in the range.

Since the LHS is homogenous, we can let $a+b+c=1$ .But note that $\sqrt{\frac{a}{1-a}}\geq 2a$ , with equality exactly when a=1/2 or 0.So $F(a,b,c)=\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\geq 2.a+2.b+2.c=2$ , with equality when two if them are each a half and the other is zero.But since a,b,c>0, the inequality is strong, meaning that we can get as close as we want to 2, but never reach it.

$\displaystyle \sum \sqrt{\dfrac{a}{b+c}}=\displaystyle \sum \dfrac{a}{\sqrt{a(b+c)}}\stackrel{\text{AM-GM}}{\geq} \displaystyle \sum \dfrac{2a}{a+(b+c)}=2$

Equality cases are when two variables are the same and one is zero (or, since the question requieres positivity, approaches zero)

First observe that if $0 < x < n$ , then $\sqrt{\frac{x}{n-x}} \ge \frac{2}{n} x$ . The proof of this is simple algebra - this equation is equivalent to $x(2x-n)^2 \ge 0$ .

Let $n = a+b+c$ . Then, $\sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}} = \sqrt{\frac{a}{n-a}} + \sqrt{\frac{b}{n-b}} + \sqrt{\frac{c}{n-c}}$ $\ge \frac{2}{n} a + \frac{2}{n} b + \frac{2}{n} c = \frac{2}{n} * n = 2$

Therefore $2$ is a lower bound on $k$ . Let's show that it is also an upper bound. Let $b = \frac{n-a}{2}$ and $c = \frac{n-a}{2}$ . Then the function becomes: $\sqrt{\frac{a}{n-a}} + \sqrt{\frac{n-a}{n+a}} + \sqrt{\frac{n-a}{n+a}}$

Hence, as $a$ approaches zero, $b$ and $c$ approach $\frac{n}{2}$ , and the function approaches $2$ . Hence it can get arbitrarily close to $2$ , and therefore the minimum value is at most 2. Hence $k \le 2$ .

Therefore, we must have $k = \boxed{2}$ .

Normally, equality occurs when $a=b=c$ . A quick substitution to the problem we get $k=\frac{3}{\sqrt2}$ .

However, what if one of the variable gets $0$ ? Say, $a=0$ .

$\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}\geq k$

Now the question gets elementary, which answer to the question is $k=2$ .

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Assuming $a=b=c$ will normally get us to the minimum value of an equation. This is because it is the equality case of AM-GM Inequality. However, AM-GM can only be applied to a set of positive real numbers. They ignore the fact that a variable can be $0$ . So , when we let $a=0$ , a minimum value occurs.

Noticed that $a,b,c$ are positive triples of real number. So assume

$\displaystyle a=\lim_{n \to 0}n$

I believe this is the reason Calvin wants $k$ to 2 decimal places. (Because it is not exactly $2$ )

Sample: a = 0.001, b = 326.058 and c = 329.825 for 2.00126471003224; the largest lower limit ought to be 2.00 by observation.

PROGRAM abc; {2.00}

USES CRT;

VAR

a, b, c, huge, range: LONGINT;

a , b , c_, k, p, am, bm, cm: EXTENDED;

BEGIN

 CLRSCR;

 RANDOMIZE;

 huge:=0; {1, 10, 100 and etc would not produce minimum.}

 range:=huge+10;

 k:=3.0*SQRT(0.5);

 REPEAT

       FOR a:=huge TO range DO

           FOR b:=huge TO range DO

               FOR c:=huge TO range DO

               BEGIN  {Allow occasional division by zero for efficiency.}

                    a_:=0.001*RANDOM(1000)+a;

                    b_:=0.001*RANDOM(1000)+b;

                    c_:=0.001*RANDOM(1000)+c;

                    p:=SQRT(a_/(b_+c_))+SQRT(b_/(c_+a_))+SQRT(c_/(a_+b_));

                    IF (p<k) AND (a_<>0.0) AND (b_<>0.0) AND (c_<>0.0) THEN

                    BEGIN

                       k:=p;

                       am:=a_;

                       bm:=b_;

                       cm:=c_

                    END

               END;

       WRITELN(k:1:18, '    ', range, '    ', am:1:3, '    ', bm:1:3, '    ', cm:1:3);

       range:=range+10

 UNTIL KEYPRESSED;

 WRITE(k:1:18);

 READLN

END.

[This is not a solution.]

For a hint, see this problem .

For those who can solve it, you will realize that the equality condition does not occur when $a = b = c$ , hence the title. I would appreciate if you can contribute to the Wiki page Inequality with strange equality conditions .