2 replies to this topic

[BysLyl]

Cho $a,b,c\in \begin{bmatrix} 0;1 \end{bmatrix}$ . Chứng minh:

$\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq 2$

[Christian Goldbach]

Không mất tính TQ gs $a\geq b\geq c$

Khi đó:

$\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq \frac{a+b+c}{bc+1}$

Ta cm $a+b+c\leq 2(bc+1)$

Thực vậy $(b-1)(c-1)\geq 0\Rightarrow bc+1\geq b+c;1\geq a;bc\geq 0\Rightarrow 2(bc+1)\geq a+b+c$

Vậy bđt được cm

[Vu Thuy Linh]

Cho $a,b,c\in \begin{bmatrix} 0;1 \end{bmatrix}$ . Chứng minh:

$\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq 2$

Giả sử $a\geq b\geq c$ => $\frac{b}{ac+1}+\frac{c}{ab+1}\leq \frac{b}{bc+1}+\frac{c}{bc+1}=\frac{b+c}{bc+1}$

Vì $0\leq b,c\leq 1=>(1-b)(1-c)\geq 0\Leftrightarrow bc+1\geq b+c$

$\Rightarrow \frac{b+c}{bc+1}\leq 1$

Mà $0\leq a,b,c\leq 1\Rightarrow a\leq 1\leq 1+bc\Rightarrow \frac{a}{bc+1}\leq 1$ => VT $\leq 2$