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

Theo tính chất tỉ lệ thức

\(\frac{x^2+y^2}{a+b}=\frac{x^2}{a}=\frac{y^2}{b}\left(a;b\ne0\right)\)

\(\frac{x^{2012}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}=\left(\frac{x^2}{a}\right)^{1006}+\left(\frac{y^2}{b}\right)^{1006}=2.\left(\frac{x^2+y^2}{a+b}\right)^{2006}=\frac{2}{\left(a+b\right)^{2006}}\left(đpcm\right)\)

đề phải ntn chứ \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\)

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}\ge\dfrac{\left(x^2+y^2\right)^2}{a+b}=\dfrac{1}{a+b}\)(cauchy-schwarz)

dấu = xảy ra khi \(\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow bx^2=ay^2\)

Xét \(\left(x^{2012}+y^{2012}\right)-\left(x^{2011}+y^{2011}\right)\)

\(=x^{2011}\left(x-1\right)+y^{2011}\left(y-1\right)\)

\(=x^{2011}\left(1-y\right)+y^{2011}\left(y-1\right)\) (do \(x-1=1-y\))

\(\Leftrightarrow\left(x^{2012}+y^{2012}\right)-\left(x^{2011}+y^{2011}\right)=\left(1-y\right)\left(x^{2011}-y^{2011}\right)\)

+ Giả sử \(x\ge y\Rightarrow x^{2011}\ge y^{2011}\) và \(x\ge1\ge y\)

Do đó \(\left(1-y\right)\left(x^{2011}-y^{2011}\right)\ge0\) (Đpcm)

+ Tương tự nếu \(y\ge x\Rightarrow y^{2011}\ge x^{2011}\) và \(y\ge1\ge x\)

Do đó \(\left(1-y\right)\left(x^{2011}-y^{2011}\right)\ge0\) (Đpcm)

Dấu "=" xảy ra khi \(x=y=1\)

Xét \left(x^{2012}+y^{2012}\right)-\left(x^{2011}+y^{2011}\right)(x2012+y2012)−(x2011+y2011)

=x^{2011}\left(x-1\right)+y^{2011}\left(y-1\right)=x2011(x−1)+y2011(y−1)

=x^{2011}\left(1-y\right)+y^{2011}\left(y-1\right)=x2011(1−y)+y2011(y−1) (do $x-1=1-y$)

\Leftrightarrow\left(x^{2012}+y^{2012}\right)-\left(x^{2011}+y^{2011}\right)=\left(1-y\right)\left(x^{2011}-y^{2011}\right)⇔(x2012+y2012)−(x2011+y2011)=(1−y)(x2011−y2011)

+ Giả sử x\ge y\Rightarrow x^{2011}\ge y^{2011}x≥y⇒x2011≥y2011 và $x\ge 1\ge y$

Do đó \left(1-y\right)\left(x^{2011}-y^{2011}\right)\ge0(1−y)(x2011−y2011)≥0 (Đpcm)

+ Tương tự nếu y\ge x\Rightarrow y^{2011}\ge x^{2011}y≥x⇒y2011≥x2011 và $y\ge 1\ge x$

Do đó $($