Giải thích các bước giải:

Ta có:

$S_1-S_2=(\dfrac{a^2}{a+b}-\dfrac{b^2}{a+b})+(\dfrac{b^2}{b+c}-\dfrac{c^2}{b+c})+(\dfrac{c^2}{c+a}-\dfrac{a^2}{c+a})$ 

$\rightarrow S_1-S_2=\dfrac{a^2-b^2}{a+b}+\dfrac{b^2-c^2}{b+c}+\dfrac{c^2-a^2}{c+a}$ 

$\rightarrow S_1-S_2=\dfrac{(a-b)(a+b)}{a+b}+\dfrac{(b-c)(b+c)}{b+c}+\dfrac{(c-a)(c+a)}{c+a}$ 

$\rightarrow S_1-S_2=a-b+b-c+c-a=0$

$\rightarrow S_1=S_2\rightarrow đpcm$

$\rightarrow 2S_1=(\dfrac{a^2}{a+b}+\dfrac{b^2}{a+b})+(\dfrac{b^2}{b+c}+\dfrac{c^2}{b+c})+(\dfrac{c^2}{c+a}+\dfrac{a^2}{c+a})$ 

$\rightarrow 2S_1=\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}$ 

$\rightarrow 2S_1\ge \dfrac{(a+b)^2/2}{a+b}+\dfrac{(b+c)^2/2}{b+c}+\dfrac{(c+a)^2/2}{c+a}$ 

$\rightarrow 2S_1\ge \dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}$ 

$\rightarrow 2S_1\ge a+b+c$

$\rightarrow S_1\ge \dfrac{a+b+c}{2}$