Đặt \(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\)
Ta có
\(\dfrac{a^2}{b+c}=a\left(\dfrac{a}{b+c}\right)=a\left(\dfrac{a+b+c-\left(b+c\right)}{b+c}\right)=\dfrac{a\left(a+b+c\right)}{b+c}-a\)
Tương tự \(\dfrac{b^2}{a+c}=\dfrac{b\left(a+b+c\right)}{a+c}-b\) ; \(\dfrac{c^2}{a+b}=\dfrac{c\left(a+b+c\right)}{a+b}-c\)
Cộng vế với vế:
\(A=\dfrac{a}{b+c}\left(a+b+c\right)+\dfrac{b}{a+c}\left(a+b+c\right)+\dfrac{c}{a+b}\left(a+b+c\right)-\left(a+b+c\right)\)
\(A=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)-\left(a+b+c\right)\)
\(A=a+b+c-\left(a+b+c\right)=0\) (đpcm)
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(=a\left(\dfrac{a}{b+c}+1\right)+b\left(\dfrac{b}{c+a}+1\right)+c\left(\dfrac{c}{a+b}+1\right)-\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-\left(a+b+c\right)\)
\(=\left(a+b+c\right).1-\left(a+b+c\right)=0\left(đpcm\right)\)
Giả thiết: a/(b+c) + b/(c+a) + c/(a+b) = 1
Đặt A = a²/(b+c) + b²/(c+a) + c²/(a+b) = a[a/(b+c)] + b[b/(c+a)] + c[c/(a+b)]
A = a[a/(b+c) + 1 - 1] + b[b/(c+a) + 1 - 1] + c[c/(a+b) + 1 - 1]
A = a.(a+b+c)/(b+c) -a + b.(a+b+c)/(c+a) - b + c.(a+b+c)/(a+b) - c
A = (a+b+c)[a/(b+c) + b/(c+a) + c/(a+b)] - (a+b+c)
A = (a+b+c) - (a+b+c) = 0
