AM-GM Problem Limbo

Here's one of the way other than using AM-GM .

Let $a \geq b \geq c,$ so $\dfrac{1}{b+c}\geq\dfrac{1}{a+c}\geq \dfrac{1}{a+b}.$

Now using the Rearrangement Inequality , we have

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\geq\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{a}{b+a}..........(1)$

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\geq\dfrac{c}{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+a}..........(2)$

Add $(1)$ and $(2)$ to get

$2(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a})\geq\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}+\dfrac{a+b}{b+a}$

$2(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a})\geq3$

$(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a})\geq \boxed{\dfrac{3}{2}}$

with equality occurring where $a=b=c.$

By Cauchy-Schwarz ,

$((a+b)+(b+c)+(c+a))(\frac {1}{a+b}+\frac {1}{b+c}+\frac {1}{c+a}) \geq 9$ . Dividing by 2 and subtracting 3 gives the minimum of 1.5 when $a=b=c$ .

Let $S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$

We know that (by AM-GM):

$\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \ge 3$ $(\frac{a}{b+c})+\frac{b}{b+c}+(\frac{b}{c+a})+\frac{c}{c+a}+(\frac{c}{a+b})+\frac{a}{a+b} \ge 3$ $\Rightarrow S+\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b} \ge 3$ $\Rightarrow S+(1-\frac{c}{b+c})+(1-\frac{a}{c+a})+(1-\frac{b}{a+b}) \ge 3$ $\Rightarrow S-(\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}) \ge 0$ $\Rightarrow S \ge \frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$ $\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b} \boxed{I}$

The minimum occurs when $LHS = RHS$ . We see that the equality case of $\boxed{I}$ is a cyclic equation, from which we deduce that $a=b=c$

Therefore, the minimum value is:

$\boxed{min(S) = \frac{3}{2} }$

Nesbitt's inequality... where k=3/2.

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3\geq \frac{9}{2}$ $\Leftrightarrow 2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq 9$ $\Leftrightarrow (a+b+b+c+c+a)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq 9$ (Always happen)

by AM-GM,

$\frac{a}{b+c}$ + $\frac{b}{c+a}$ + $\frac{c}{a+b}$ $\geq$ 3 $\times$ $\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}$

The expression is at minimum when a = b = c, therefore 3 $\times$ $\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}$ = $3\times$ $\sqrt[3]{\frac{a^3}{(2a)(2a)(2a)}}$ = $3\times\frac{1}{2}$ = 1.5

Starting with $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ , if we add and subtract one from each term, we get

$\frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} + \frac{a+b+c}{a+b} - 3$

$=(a+b+c)(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}) - 3$

$=0.5((b+c)+(c+a)+(a+b))(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}) - 3$

now applying Cauchy-Schwarz , we get

$≥0.5(3^2) - 3$

$≥\frac{3}{2}$

Thus, the minimum of the expression is $\frac{3}{2}$ .

Let minimum be 1 put get 1.5