toán chứng minh!khó đó nhe!

V

vy000

Với số dương $\dfrac xy <1$ ta có: $\dfrac xy< \dfrac{x+n}{y+n}(n>0)$

Áp dụng:

$\dfrac a{a+b+c}+\dfrac b{b+c+d}+\dfrac c{c+d+a}+\dfrac d{d+a+b} < \dfrac {a+d}{a+b+c+d}+\dfrac {b+a}{a+b+c+d}+\dfrac {c+b}{a+b+c+d}+\dfrac {d+c}{a+b+c+d}=2$
 
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