Easy inequality

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} = \dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{bc+ac}$

Using Titu's Lemma ,

$\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{bc+ac} \geq \dfrac{(a+b+c)^2}{2(ab+bc+ac)} \geq \boxed{\dfrac{3}{2}}$

Since $(a+b+c)^2 \geq 3(ab+bc+ac)$ .

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Proof for: $(a+b+c)^2 \geq 3(ab+bc+ac)$ .

The above inequality is equivalent to $a^2 + b^2 + c^2 \ge ab + bc + ca$ which can be easily established by trivial inequality $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (a - c)^2] \ge 0$ .

$\large\mathfrak{T}= \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$ $=\dfrac{1}{\color{#3D99F6}{2}}\left(\dfrac{\color{#3D99F6}{2}a}{b+c}+\dfrac{\color{#3D99F6}{2}b}{c+a}+\dfrac{\color{#3D99F6}{2}c}{a+b}\right)$ $=\dfrac 12\left(\color{#D61F06}{1+}\dfrac{2a}{b+c}+\color{#D61F06}{1+}\dfrac{2b}{c+a}+\color{#D61F06}{1+}\dfrac{2c}{a+b} \color{#D61F06}{-3}\right)$ $=\dfrac 12\underbrace{\left(\dfrac{a+b}{b+c}+\dfrac{a+c}{b+c}+\dfrac{b+c}{c+a}+\dfrac{b+a}{c+a}+\dfrac{a+c}{a+b}+\dfrac{b+c}{a+b}\right)}_{\color{#456461}{\text{Applying }AM\geq GM}}-\dfrac 32$ $\large\mathfrak{T}\geq \dfrac 12\left(6\right)-\dfrac{3}{2}=\dfrac 32$ $\Huge\therefore 2+3=\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{5}}}}}$ $\boxed{\color{forestgreen}{\mathbf{NOTE:-}\\ \text{For Equality }a=b=c\neq 0}}$

Solution by only using AM-GM inequality,

$\begin{aligned} S & = \frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b} \\ A & = \frac{b}{b+c}+ \frac{c}{c+a}+ \frac{a}{a+b} \\ B & = \frac{c}{b+c}+ \frac{a}{c+a}+ \frac{b}{a+b} \\ \therefore A+B &=3 \\ S+A &= \frac{a+b}{b+c}+ \frac{b+c}{c+a}+ \frac{c+a}{a+b} ≥ 3 \\ S+B &= \frac{a+c}{b+c}+ \frac{b+a}{c+a}+ \frac{c+b}{a+b} ≥ 3 \\ A+B+2S & \geq 6\Rightarrow S \geq \frac{3}{2} \\ \Rightarrow 3+2 & = \boxed{5} \end{aligned}$

Here's my solution:

Let's normalize by assuming $a+b+c=1$ . The above expression reduces to $\sum_{cyc} \frac{a}{1-a}$ now, let $f(x)=\frac{x}{1-x}$ . Since $f(x)$ is a convex function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ so, we result $f(a)+f(b)+f(c) \leq \frac{3}{2}$ . So, $\frac{3}{2}=\frac{a}{b}$ and hence, the value of $a+b$ is $3+2=\boxed{5}$