Đề bài: Cho $a,b,c,d>0$.Chứng minh rằng:$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}$

Lời giải

Đề bài:
Cho $a,b,c,d>0$.Chứng minh rằng:$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}$
Lời giải

Đặt:
$\begin{cases}6x=b+2c+3d >0\\ 6y= c+2d+3a >0\\ 6z=d+2a+3b>0 \\ 6t=a+2b+3c>0\end{cases}$
$\Leftrightarrow \begin{cases}4a=-5x+7y+z+t \\ 4b=x-5y+7z
+t\\ 4c=x+y-5z+7t\\ 4d=7x+y+z-5t\end{cases}$
Theo BĐT Cauchy:
$4(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}+\frac{d}{t})=\frac{-5x+7y+z+t}{x}+\frac{x-5y+7z
+t}{y}+\frac{x+y-5z+7t}{z}+\frac{7x+y+z-5t}{t}$
$=-20+7(\frac{y}{x}+\frac{z}{y}+\frac{t}{z}+\frac{x}{t})+(\frac{x}{z}+\frac{z}{x})+(\frac{y}{t}+\frac{t}{y})+(\frac{t}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{t})$
$\geq -20+7.4+2+2+4=16$
$\Rightarrow \frac{a}{x}+\frac{b}{y}+\frac{c}{z}+\frac{d}{t}\geq 4$
$\Rightarrow \frac{a}{6x}+\frac{b}{6y}+\frac{c}{6z}+\frac{d}{6t}\geq \frac{4}{6}=\frac{2}{3}$
Dấu “=” xảy ra $\Leftrightarrow x=y=z=t \Leftrightarrow a=b=c=d$
$\Rightarrow $ (ĐPCM)

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Chuyên mục: Bất đẳng thức Côsi