Lời giải
Đề bài:
Cho $0\leq a,b,c,d\leq 1$.Chứng minh rằng:$\frac{a}{bcd+1}+\frac{b}{cda+1}+\frac{c}{dab+1}+\frac{d}{abc+1}\leq 3$
Lời giải
Ta có: $\left ( 1-a \right )\left ( 1-b \right )+\left ( 1-c \right )\left ( 1-d \right )+\left ( 1-ab \right )\left ( 1-cd \right )\geq 0$
$\Rightarrow 1-\left ( a+b \right )+ab+1-\left ( c+d \right )+cd+1-\left ( ab+cd \right )+abcd\geq 0$
$\Rightarrow 3+abcd\geq a+b+c+d$
Suy ra:
$\frac{a}{bcd+1}+\frac{b}{cda+1}+\frac{c}{dab+1}+\frac{d}{abc+1}\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=$
$=\frac{a+b+c+d}{abcd+1}\leq \frac{3+abcd}{abcd+1}\leq \frac{3+abcd}{abcd+1}\leq 3$( vì $ abcd\geq 0$)
$\Rightarrow $ ĐPCM
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