Question

[giang nguyễn]

cho 3 số dương 0<a<b<c<1 cmr:\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}< 2\)

Answer 1

[Akai Haruma]

Lời giải:

Do $0< a< b< c< 1$ nên $0< ab< ac< bc$

\(\Rightarrow \frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}< \frac{a}{ab+1}+\frac{b}{ab+1}+\frac{c}{ab+1}=\frac{a+b+c}{ab+1}(1)\)

Vì $a,b< 1$ nên \((a-1)(b-1)>0\Leftrightarrow ab+1> a+b\)

$c< 1$ nên $1+ab>c$

\(\Rightarrow 2(ab+1)> a+b+c(2)\)

Từ (1);(2) \(\Rightarrow \frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}< \frac{a+b+c}{ab+1}< \frac{2(ab+1)}{ab+1}=2\)

Ta có đpcm.