) gt: a/(b+c) + b/(c+a) + c/(a+b) = 1
A = a²/(b+c) + b²/(c+a) + c²/(a+b) = a[a/(b+c)] + b[b/(c+a)] + c[c/(a+b)]
= a[a/(b+c) + 1 - 1] + b[b/(c+a) + 1 - 1] + c[c/(a+b) + 1 - 1]
= a.(a+b+c)/(b+c) -a + b.(a+b+c)/(c+a) - b + c.(a+b+c)/(a+b) - c
= (a+b+c)[a/(b+c) + b/(c+a) + c/(a+b)] - (a+b+c)
= (a+b+c) - (a+b+c) = 0
Ta có : \(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Rightarrow\frac{\left(a+b+c\right)a}{b+c}+\frac{\left(a+b+c\right)b}{c+a}+\frac{\left(a+b+c\right)c}{a+b}=a+b+c\)
\(\Rightarrow\frac{a^2+ab+ac}{b+c}+\frac{ab+b^2+bc}{c+a}+\frac{ac+bc+c^2}{a+b}=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{ab+ac}{b+c}+\frac{b^2}{a+c}+\frac{ab+bc}{c+a}+\frac{c^2}{a+b}+\frac{ac+bc}{a+b}=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+a+b+c-a-b-c=0\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\left(đpcm\right)\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=a.\left(\frac{a}{b+c}\right)+b.\left(\frac{b}{c+a}\right)+c.\left(\frac{c}{a+b}\right)\)
\(=a\left[\frac{a}{\left(b+c\right)}+1-1\right]+b\left[\frac{b}{c+a}+-1\right]+c\left[\frac{c}{a+b}+1-1\right]\)
\(=\frac{a\left(a+b+c\right)}{\left(b+c\right)}-a+\frac{b\left(a+b+c\right)}{\left(c+a\right)}-b+\frac{c\left(a+b+c\right)}{\left(a+b\right)}-c\)
\(=\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)}+\frac{b}{\left(a+c\right)}+\frac{c}{a+b}\right]-\left(a+b+c\right)\)
\(\left(a+b+c\right)-\left(a+b+c\right)=0\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)
