Chủ đề này có 2 trả lời

[axe900]

Cho a,b,c,d > 0 thoa man $\frac{1}{1+a}$+$\frac{1}{1+b}$+$\frac{1}{1+c}$+$\frac{1}{1+d}$$\geq 3$.
CMR: $abcd\leq \frac{1}{81}$

[henry0905]

$\Rightarrow \frac{1}{a+1}\geq 1-\frac{1}{b+1}+1-\frac{1}{c+1}+1-\frac{1}{d+1}\geq 3\sqrt[3]{\frac{dbc}{(1+d)(1+b)(1+c)}}$
C/m tượng tự
$\Rightarrow \frac{1}{a+1}.\frac{1}{b+1}.\frac{1}{c+1}.\frac{1}{d+1}\geq 81\frac{abcd}{(1+a)(1+b)(1+c)(1+d)}$
$\Rightarrow abcd\leq \frac{1}{81}$ (dpcm)

Bài viết đã được chỉnh sửa nội dung bởi henry0905: 01-07-2012 - 11:33

[nucnt772]

ta có: $\frac{1}{1+a}$+$\frac{1}{1+b}$+$\frac{1}{1+c}$+$\frac{1}{1+d}\geq 3$

$\Rightarrow$ $(1-\frac{1}{1+a})$+$(1-\frac{1}{1+b})$+$(1-\frac{1}{1+c})$+$(1-\frac{1}{1+d})\leq 1$

$\Leftrightarrow$ $\frac{a}{1+a}$+$\frac{b}{1+b}$+$\frac{c}{1+c}$+$\frac{d}{1+d}\leq 1$

$\Leftrightarrow$ $[(1+a)+(1+b)+(1+c)+(1+d)].$$(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d})$$\leq 4+a+b+c+d$ (1)

áp dụng BĐT Bunhiacopxki cho 8 số: $\sqrt{1+a}$, $\sqrt{1+b}$, $\sqrt{1+c}$, $\sqrt{1+d}$, $\frac{\sqrt{a}}{1+a}$, $\frac{\sqrt{b}}{1+b}$, $\frac{\sqrt{c}}{1+c}$, $\frac{\sqrt{d}}{1+d}$ :

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

từ (1) và (2) $\Rightarrow$ $(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^{2}$$\leq 4+a+b+c+d$

$\Leftrightarrow$ $(\sqrt{ab}+\sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}+\sqrt{cd})\leq 2$

áp dụng BĐT Cauchy cho 6 số ta được:

$2\geq 6.\sqrt[6]{\sqrt{a^{3}.b^{3}.c^{3}.d^{3}}}$

$\Leftrightarrow \sqrt[4]{abcd}\leq \frac{1}{3}$ $\Rightarrow abcd\leq \frac{1}{81}$ (đpcm)