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Bất đẳng thức cần chứng minh tương đương với

$\frac{a}{d+b}+\frac{b}{a+c}+\frac{c}{a+d}+\frac{d}{b+c}\ge \frac{a}{a+d}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{d}{d+b}$

 

$\Leftrightarrow \frac{a+b}{d+b}+\frac{b+a}{a+c}+\frac{c+d}{a+d}+\frac{d+c}{b+c}\ge 4$

 

$\Leftrightarrow (a+b)\left(\frac{1}{d+b}+\frac{1}{a+c}\right)+(c+d)\left(\frac{1}{a+d}+\frac{1}{b+c}\right)\ge 4$

 

Áp dụng BĐT $\frac{1}{x}+\frac{1}{y}\ge \frac{4}{x+y}$, ta có:

 

$(a+b)\left(\frac{1}{d+b}+\frac{1}{a+c}\right)+(c+d)\left(\frac{1}{a+d}+\frac{1}{b+c}\right)\ge\frac{4(a+b)}{d+b+a+c}+\frac{4(c+d)}{a+d+b+c}=4$

làm theo hướng này nè hoanghung9g

BĐT Cauchy-Schwarz chứng minh sao ạ 

Ta có: $\frac{1}{a+1}\geq 2-(\frac{1}{b+1}+\frac{1}{c+1})=(1-\frac{1}{b+1})+(1-\frac{1}{c+1})=\frac{b}{b+1}+\frac{c}{c+1}\geq$$2\times \sqrt{\frac{bc}{(b+1)(c+1)}}$ (1)

CMTT: $\frac{1}{b+1}\geq 2\times \sqrt{\frac{ac}{(a+1)(c+1)}}$ (2)

$\frac{1}{c+1}\geq 2\times \sqrt{\frac{ab}{(a+1)(b+1)}}$ (3)

Lấy (1)$\times$(2)$\times$(3) , ta được: $\frac{1}{(a+1)(b+1)(c+1)}\geq \frac{8abc}{(a+1)(b+1)(c+1)}$

$\Leftrightarrow 8abc\leq 1\Leftrightarrow abc\leq \frac{1}{8}$

Dấu "=" xr $\Leftrightarrow a=b=c=\frac{1}{2}$

$\Rightarrow max P=\frac{1}{8}\Leftrightarrow a=b=c=\frac{1}{2}$