cho a,b,c >0.Chứng minh rằng

K

khaiproqn81

a) Ta có: $(a-b)^2 \ge 0 \\ \to a^2+b^2-2ab \ge 0 \\ \to \dfrac{a^2+b^2}{ab} \ge 2 \\ \to \dfrac{a}{b}+\dfrac{b}{a} \ge 2$
 
K

khaiproqn81

c) $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=1+1+1+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a} \ge 3+2+2+2=9$
 
C

congchuaanhsang

b, (a+b)($\dfrac{1}{a}+\dfrac{1}{b}$)\geq4
c, (a+b+c)($\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$)\geq9
Bấm để xem đầy đủ nội dung ...

b, Cauchy-Schwarz:

$(a+b) (\dfrac{1}{a}+\dfrac{1}{b}) \ge (a+b)\dfrac{4}{a+b}=4$

c, Cauchy-Schwarz:

$(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}) \ge (a+b+c)\dfrac{9}{a+b+c}=9$

Chị cứ chậm chậm:)) , làm gì mak sai $Latex$ thế =))

Vội quá =))
 
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C

congchuaanhsang

b, (a+b)($\dfrac{1}{a}+\dfrac{1}{b}$)\ge4
c, (a+b+c)($\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$)\geq9
Bấm để xem đầy đủ nội dung ...

b, Cauchy $(a+b)(\dfrac{1}{a}+\dfrac{1}{b}) \ge 2\sqrt{ab}.\dfrac{2}{\sqrt{ab}}=4$

c, Cauchy $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$

\geq $3\sqrt[3]{abc}.\dfrac{3}{\sqrt[3]{abc}}=9$

Đã bảo là chậm chậm, ko ai giựt đâu, người gì xấu tính =))
 
T

transformers123

còn cách dùng bđt Bunyakovsky sao ko ai làm vậy=))
b/ ta có:
$(a+b)(\dfrac{1}{a}+\dfrac{1}{b}) \ge (\dfrac{\sqrt{a}}{\sqrt{a}}+\dfrac{\sqrt{b}}{\sqrt{b}})^2=4$
c/ áp dụng tiếp:
$(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}) \ge (\dfrac{\sqrt{a}}{\sqrt{a}}+\dfrac{\sqrt{b}}{\sqrt{b}}+\dfrac{\sqrt{c}}{\sqrt{c}})^2=9$
 
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