Cho các số thực dương $a, b, c$ thỏa mãn $a \leq b \leq c$ và $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
Tìm min $P=a.b^{2}.c^{3}$
$\frac 1a+\frac 2b+\frac 3c \overset{AM-GM}{\ge} \frac 6{\sqrt[6]{ab^2c^3}}$
$\Leftrightarrow \sqrt[6]{ab^2c^3} \ge \frac{6}{\dfrac 1a+\dfrac 2b+\dfrac 3c}$ (1)
Lại có 
$\frac 1a+\frac 2b+\frac 3c = \left( \frac 1a-1 \right) + \left(  \frac 2b -2\right)+\left( \frac 3c-3 \right)+6$
$=\frac{1-a}{a}+\frac{2(1-b)}{b}+\frac{3(1-c)}{c}+6  \le \frac{1-a}{a}+\frac{2(1-b)}{a}+\frac{3(1-c)}{a}+6$
$=\frac{6-(a+2b+3c)}{a}+6 \le \frac{6-2(a+b+c)}{a}+6$
$=\frac{6-2\sqrt{(a+b+c)(\frac 1a+\frac 1b+\frac 1c)}}{a}+6 \le \frac{6-2.\sqrt 9}{a}+6=6$ (2)
~~~~~~~~~~
Từ (1) & (2) $\Rightarrow \sqrt[6]{ab^2c^3} \ge 1\Rightarrow P \ge 1$
Vậy $\min P=1\Leftrightarrow a=b=c=1$
a sửa rồi đó, chỉ nhầm cái dấu mà ko nhận ra à -_- –  tran85295 12-05-16 01:25 PM
đáng lẽ tới đó e phải biết rồi chứ –  tran85295 12-05-16 01:22 PM
hehe :))) –  tran85295 12-05-16 01:21 PM
giải thích cái coi –  Ngọc 12-05-16 01:20 PM
ý em là sao anh đánh giá P< hoặc= 1 mà biểu là min? –  Ngọc 12-05-16 01:20 PM
dấu má bị sao zậy? –  Ngọc 12-05-16 01:09 PM
ý e là sao đánh giá =< mà kêu là min? –  Ngọc 12-05-16 01:08 PM
e nhìn là biết :3 –  tran85295 12-05-16 11:39 AM
nè,anh Nam,đánh giá kiểu j kì vậy? –  Ngọc 12-05-16 11:23 AM

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