Chứng minh bất đẳng thức (2)

E

eye_smile

1,Bài này có nhiều nhiều cách
Tham khảo 1 cách nhá:
$\dfrac{{a^3}}{b}+ab$ \geq $2{a^2}$
$\dfrac{{b^3}}{c}+bc$ \geq $2{b^2}$
$\dfrac{{c^3}}{a}+ca$ \geq $2{c^2}$
Cộng theo vế, được: $A+ab+bc+ca$ \geq $2{a^2}+2{b^2}+2{c^2}$ \geq 2ab+2bc+2ca
\Leftrightarrow A \geq ab+bc+ca
-->đpcm
 
V

vipboycodon

Câu 1:
Theo cauchy-schwarz ta có:
$\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} = \dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac} \ge \dfrac{(a^2+b^2+c^2)^2}{ab+bc+ac} $
Mặt khác: $a^2+b^2+c^2 \ge ab+bc+ac$ (cauchy)
=> đpcm
Dấu "=" xảy ra khi $a = b = c$.
 
C

congchuaanhsang

2, Xét $|\dfrac{(a+b)(1-ab)}{(a^2+1)(b^2+1)}|$=$|\dfrac{(a+b)(1-ab)}{(a^2+2ab+b^2)+(a^2b^2-2ab+1)}|$

=$|\dfrac{(a+b)(1-ab)}{(a+b)^2+(1-ab)^2}|$=$\dfrac{|(a+b)(1-ab)|}{(a+b)^2+(1-ab)^2}$

\leq$\dfrac{|a+b|^2+|1-ab|^2}{2[(a+b)^2+(1-ab)^2]}$=$\dfrac{(a+b)^2+(1-ab)^2}{2[(a+b)^2+(1-ab)^2]}$=$\dfrac{1}{2}$

\Leftrightarrow đpcm
 
T

trungthinh.99

congchuaanhsang said:
2, Xét $|\dfrac{(a+b)(1-ab)}{(a^2+1)(b^2+1)}|$=$|\dfrac{(a+b)(1-ab)}{(a^2+2ab+b^2)+(a^2b^2-2ab+1)}|$

=$|\dfrac{(a+b)(1-ab)}{(a+b)^2+(1-ab)^2}|$=$\dfrac{|(a+b)(1-ab)|}{(a+b)^2+(1-ab)^2}$

\leq$\dfrac{|a+b|^2+|1-ab|^2}{2[(a+b)^2+(1-ab)^2]}$=$\dfrac{(a+b)^2+(1-ab)^2}{2[(a+b)^2+(1-ab)^2]}$=$\dfrac{1}{2}$

\Leftrightarrow đpcm
Bấm để xem đầy đủ nội dung ...

Còn trường hợp $\frac{-1}{2}$ \leq nữa bạn ơi. :D

____________________________________________________
@congchua |A|\leq$\dfrac{1}{2}$\Leftrightarrow$\dfrac{-1}{2}$\leqA\leq$\dfrac{1}{2}$ !
 
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E

eye_smile

Bạn ấy đã c/m được $|A|$ \leq $\dfrac{1}{2}$
\Rightarrow $\dfrac{-1}{2}$ \leq A \leq $\dfrac{1}{2}$
 
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