Áp dụng bđt Cauchy Shwarz dạng Engel, ta có:
\(\dfrac{x^4}{a}+\dfrac{y^4}{b}\ge\dfrac{\left(x^2+y^2\right)^2}{a+b}=\dfrac{1}{a+b}\) (vì \(x^2+y^2=1\))
mà \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\) (theo đề bài)
\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\) (tính chất của dãy tỉ số bằng nhau)
\(\Rightarrow x^2=\dfrac{a}{a+b}\)
\(B=\dfrac{x^{2008}}{a^{1004}}+\dfrac{y^{2008}}{b^{1004}}\)
\(=\left(\dfrac{x^2}{a}\right)^{1004}+\left(\dfrac{y^2}{b}\right)^{1004}\)
\(=2\times\left(\dfrac{\dfrac{a}{a+b}}{a}\right)^{1004}\) (vì \(\dfrac{x^2}{a}=\dfrac{y^2}{b}\))
Thay số vào ròi tính thoy ~~! (xxx)
\(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{\left(x^2+y^2\right)^2}{a+b}\left(dox^2+y^2=1\right)\)
\(\Leftrightarrow\dfrac{x^4}{a}+\dfrac{y^4}{b}-\dfrac{\left(x^2+y^2\right)^2}{a+b}=0\)
Tự biến hóa, hô phép ;v
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)
\(\Rightarrow\dfrac{x^{2008}}{a^{1004}}=\dfrac{y^{2008}}{b^{2004}}\Rightarrow B=2.\dfrac{x^{2008}}{a^{1004}}\)
\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\Rightarrow\dfrac{x^{2008}}{a^{1004}}=\dfrac{1}{\left(a+b\right)^{1004}}\)
\(\dfrac{2}{\left(a+b\right)^{1004}}=2.\dfrac{x^{2008}}{a^{1004}}=B\)
Vậy: \(B=\dfrac{2}{\left(a+b\right)^{1004}}\)
