Question

Với các số dương a,b,c thỏa mãn :\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

CMR: \(\frac{\left(a+b\right)^2}{c^2}+\frac{\left(c+a\right)^2}{b^2}+\frac{\left(b+c\right)^2}{a^2}=12\)

Answer 1

Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\left(a+b+c>0\right)\)

\(\Rightarrow\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=2\)

\(\Rightarrow\left(\frac{b+c}{a}\right)^2=\left(\frac{c+a}{b}\right)^2=\left(\frac{a+b}{c}\right)^2=2^2\)

\(\Rightarrow\frac{\left(b+c\right)^2}{a^2}=\frac{\left(c+a\right)^2}{b^2}=\frac{\left(a+b\right)^2}{c^2}=4\)

\(\Rightarrow\frac{\left(a+b\right)^2}{c^2}+\frac{\left(c+a\right)^2}{b^2}+\frac{\left(b+c\right)^2}{a^2}=4+4+4=12\left(đpcm\right)\)

Vậy...