[Toán 8] Chứng minh?

H

huynhbachkhoa23

Áp dụng BDT Cauchy-Schwarz:

$\dfrac{x^4}{a}+\dfrac{y^4}{b} \ge \dfrac{(x^2+y^2)^2}{a+b}=\dfrac{1}{a+b}$

$\rightarrow \dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{1}{a+b}$

$\leftrightarrow \dfrac{x^{2012}}{a^{1006}}=\dfrac{y^{2012}}{b^{1006}}=\dfrac{1}{(a+b)^{2006}}$

Suy ra điều cần chứng minh.

a, b chưa dương thì Cauchy-Schwarz thế nào được bạn
 
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R

ronaldover7

Cho [TEX]\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}[/TEX] và [TEX]x^2+y^2=1[/TEX]
Chứng minh rằng: [TEX]\frac{x^{2012}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}[/TEX]= [TEX]\frac{2}{(a+b)^{1006}[/TEX]​

[TEX]\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}[/TEX]
\Rightarrow$ \frac{x^4b+y^4a}{ab}=\frac{1}{a+b}$
\Rightarrow $(x^4b+y^4a)(a+b)=ab$
\Rightarrow $x^4b^2+y^4a^2+ab(x^4+y^4)=ab$
\Rightarrow $x^4b^2+y^4a^2+ab(1-2x^2y^2)=ab$
\Rightarrow $x^4b^2+y^4a^2-2x^2y^2ab+ab=ab$
\Rightarrow $(x^2b-y^2a)=0$
\Rightarrow$x^2b=y^2a$
\Rightarrow $\frac{x^2}{a}=\frac{y^2}{b}$
\Rightarrow $\frac{x^{2012}}{a^{1006}}=\frac{y^{2012}}{b^{1006}}$
\Rightarrow $\frac{x^{2012}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}=\frac{2x^{2012}}{a^{1006}}$

Vậy cần CM $\frac{x^{2012}}{a^{1006}}=\frac{1}{(a+b)^{1006}}$
\Rightarrow $\frac{x^2}{a}=\frac{1}{a+b}$

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