Ta có : \(\frac{2014a^2+b^2+c^2}{a^2}=\frac{a^2+2014b^2+c^2}{b^2}=\frac{a^2+b^2+2014c^2}{c^2}\)
\(\Rightarrow\) \(2014+\frac{b^2+c^2}{a^2}=2014+\frac{a^2+c^2}{b^2}=2014+\frac{a^2+b^2}{c^2}\)
\(\Rightarrow\) \(\frac{b^2+c^2}{a^2}=\frac{a^2+c^2}{b^2}=\frac{a^2+b^2}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{b^2+c^2}{a^2}=\frac{a^2+c^2}{b^2}=\frac{a^2+b^2}{c^2}=\frac{2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\) (Vì \(a^2+b^2+c^2\ne0\))
Suy ra: \(\frac{b^2}{a^2}+\frac{c^2}{a^2}=\frac{a^2}{b^2}+\frac{c^2}{b^2}=\frac{a^2}{c^2}+\frac{b^2}{c^2}=2\)
\(\Rightarrow\frac{b^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{b^2}{c^2}=2+2+2=6\)
\(\Rightarrow\) \(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}=\frac{6}{2}=3\)
Lại có: \(P=\)\(\frac{2015a^2+b^2}{c^2}+\frac{2015a^2+c^2}{b^2}+\frac{2015b^2+c^2}{a^2}\)
\(=2015\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)+\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}\right)\)
\(=\left(2015+1\right)\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)\)
\(=2016\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)\)
\(=2016.3=6048\)
Vậy \(P=6048\)
