$\sum \dfrac{\sqrt{b}}{3a+b} \le \sum \dfrac{1}{4a}$
#1
Posted 07-03-2013 - 11:35
$$\dfrac{\sqrt{b}}{3a+b}+\dfrac{\sqrt{a}}{3c+a}+\dfrac{\sqrt{c}}{3b+c} \le \dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}$$
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"If I feel unhappy,I do mathematics to become happy.
If I feel happy,I do mathematics to keep happy."
Alfréd Rényi
#2
Posted 08-03-2013 - 12:19
Bài toán: Cho $a,b,c \in \mathbb{R^+}$ thỏa mãn $a+b+c=3$.Chứng minh:
$$\dfrac{\sqrt{b}}{3a+b}+\dfrac{\sqrt{a}}{3c+a}+\dfrac{\sqrt{c}}{3b+c} \le \dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}$$
Chém gió giữa trưa
Sử dụng AM-GM ta có
$3a+b=a+a+a+b \ge 4\sqrt[4]{a^3b}$
$\Rightarrow \frac{\sqrt{b}}{3a+b}\le \frac{1}{4}\sqrt[4]{\frac{b}{a^{3}}}$
$\Rightarrow VT \le \frac{1}{4}(\sum \sqrt[4]{\frac{b}{a^{3}}})$
Cần chứng minh
$\sum \sqrt[4]{\frac{b}{a^{3}}} \le \sum \frac{1}{a}$
$\Leftrightarrow 4\sum \sqrt[4]{\frac{b}{a^{3}}} \le 4\sum \frac{1}{a}$
Lại tiếp tục sử dụng AM-GM
$4\sqrt[4]{\frac{b}{a^{3}}} \le \frac{1}{a}+\frac{1}{a}+\frac{1}{a}+b$
$\Rightarrow VT \le 3\sum \frac{1}{a}+(a+b+c)$
Cần chứng minh :
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge a+b+c=3$
Cái này lại theo C-S:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3$
$\Rightarrow$ dpcm
Dấu đẳng thức xảy ra khi $a=b=c=1$
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