Question

[Le Van Hung]

cho x,y,a,b là các số thực thỏa mãn

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^2+y^2}{a+b}\)    và   \(x^2+y^2=1\)

CM : \(\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\frac{2}{ \left( a+b\right)^{1003}}\)

Answer 1

[Nguyễn Xuân Anh]

 \(\text{Đặt }x^2=m\ge0;y^2=n\ge0\Rightarrow m+n=1\)

\(\text{Ta có: }\frac{m^2}{a}+\frac{n^2}{b}=\frac{\left(m+n\right)^2}{a+b}\Leftrightarrow\left(a+b\right)\left(\frac{m^2}{a}+\frac{n^2}{b}\right)=\left(m+n\right)^2\left(\text{BĐT Bunhiacopki}\right)\)\(\Leftrightarrow m^2+n^2+\frac{b}{a}m^2+\frac{a}{b}n^2=m^2+n^2+2mn\)

\(\Leftrightarrow\frac{b}{a}m^2+\frac{a}{b}n^2-2mn=0\left(1\right)\)

\(\text{+Nếu }\frac{a}{b}< 0\text{ thì (1)}\Leftrightarrow-\left(\sqrt{-\frac{b}{a}m}\right)^2-2mn-\left(\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}\right)^2=0\)

\(\Leftrightarrow\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}=0\Leftrightarrow m=n=0\left(\text{loại}\right)\)

\(\text{Xét }\frac{a}{b}>0;\left(1\right)\Leftrightarrow\left(\sqrt{\frac{b}{a}m}\right)^2-2mn+\left(\sqrt{\frac{a}{b}n}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}-\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\sqrt{\frac{b}{a}m}=\sqrt{\frac{a}{b}n}\)

\(\Leftrightarrow bm=an\Leftrightarrow bx^2=ay^2\left(a,b>0\right)\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\left(\frac{x^2}{a}\right)^{1003}+\left(\frac{y^2}{b}\right)^{1003}=\frac{1}{\left(a+b\right)^{1003}}+\frac{1}{\left(a+b\right)^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\left(đpcm\right)\)