x^4/a +y^4/b =1/a+b và x^2 + y^2=1
Cm: x^2020/a^1010 + y^2020/b^1010 = 2/(a+b)^1010
x^4/a +y^4/b =1/a+b và x^2 + y^2=1 Cm: x^2020/a^1010 + y^2020/b^1010 = 2/(a+b)^1010
By Mackenzie
By Mackenzie
x^4/a +y^4/b =1/a+b và x^2 + y^2=1
Cm: x^2020/a^1010 + y^2020/b^1010 = 2/(a+b)^1010
Áp dụng bất đẳng thức $Cauchy- Schwarz$ dạng $Engel$ ta được:
$\dfrac{x^4}{a} + \dfrac{y^4}{b} \geq \dfrac{(x^2 + y^2)^2}{a + b} = \dfrac{1}{a + b}$
Dấu = xảy ra $\Leftrightarrow \begin{cases}\dfrac{x^2}{a} = \dfrac{y^2}{b}\\x^2 + y^2 = 1\end{cases}$
$\Leftrightarrow \begin{cases}x^2 = \dfrac{a}{a + b}\\y^2 = \dfrac{b}{a + b}\end{cases}$
Do đó:
$\dfrac{x^{2020}}{a^{1010}} + \dfrac{y^{2020}}{b^{1010}}$
$= \left(\dfrac{x^2}{a}\right)^{1010} + \left(\dfrac{y^2}{b}\right)^{1010}$
$= 2.\left(\dfrac{x^2}{a}\right)^{1010}$
$= 2.\left(\dfrac{\dfrac{a}{a + b}}{a}\right)^{1010}$
$= 2.\left(\dfrac{1}{a + b}\right)^{1010}$
$= \dfrac{2}{(a + b)^{1010}}$
Ta có : $x^2+y^2=1$
$\to (x^2+y^2)^2 = 1^2=1$
Khi đó ta thấy $\dfrac{x^4}{a}+\dfrac{y^4}{b} = \dfrac{(x^2+y^2)^2}{a+b}$
$\to \dfrac{x^4b+y^4a}{ab} = \dfrac{x^4+y^4+2x^2y^2}{a+b}$
$\to (x^4b+y^4a).(a+b) = ab.(x^4+y^4+2x^2y^2)$
$\to x^4b^2+y^4a^2-2abx^2y^2=0$
$\to (x^2b-y^2a)^2 = 0 $
$\to x^2b=y^2a$
$\to \dfrac{x^2}{a} = \dfrac{y^2}{b} = \dfrac{x^2+y^2}{a+b} = \dfrac{1}{a+b} \to \dfrac{x^{2020}}{a^{1010}} = \dfrac{y^{2020}}{b^{1010}} =\dfrac{1}{(a+b)^{1010}}$
$\dfrac{x^{2020}}{a^{1010}} + \dfrac{y^{2020}}{b^{1010}} =\dfrac{2}{(a+b)^{1010}}$