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H

h0cmai.vn...tru0ng


$(a+b)(\frac{1}{a}+\frac{1}{b})$=$1+1+$\frac{a}{b}$+\frac{b}{a}$\geq$2+2\sqrt{\frac{ab}{ba}}$=4 .
P.S : @};-
 
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N

noinhobinhyen

Áp dụng bđt Bu-nhi-a-cốp-xki , ta có

$(\sqrt[]{a^2}+\sqrt[]{b^2})(\dfrac{1}{\sqrt[]{a^2}}+\dfrac{1}{\sqrt[]{b^2}}) $

$\geq (\sqrt[]{a}.\dfrac{1}{\sqrt[]{a}}+\sqrt[]{b}.\dfrac{1}{\sqrt[]{b}})^2$

$\Leftrightarrow (a+b)(\dfrac{1}{a}+\dfrac{1}{b}) \geq (1+1)^2=4$

Dấu [=] xảy ra khi $a=b$
 
A

asroma11235

C1:
[TEX]VT= (a+b)(\frac{1}{a}+\frac{1}{b}) \ge_{AM-GM} 2\sqrt{ab}. 2 \frac{1}{\sqrt{ab}} =4[/TEX]
C2:
[TEX](a+b)(\frac{1}{a}+\frac{1}{b}) -4 = 2+\frac{a}{b}+\frac{b}{a}-4=\frac{a}{b}+\frac{b}{a}-2=\frac{(a-b)^2}{ab} \ge 0[/TEX]
 
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