Cho $0\leq a,b,c,d\leq 1$. Chứng minh rằng:
$\frac{a}{bcd+2}+\frac{b}{cda+2}+\frac{c}{abd+2}+\frac{d}{abc+2}\leq 1+\frac{1}{abcd+2}$
Cho $0\leq a,b,c,d\leq 1$. Chứng minh rằng:
$\frac{a}{bcd+2}+\frac{b}{cda+2}+\frac{c}{abd+2}+\frac{d}{abc+2}\leq 1+\frac{1}{abcd+2}$
$$\mathbf{\text{Every saint has a past, and every sinner has a future}}.$$
Cho $0\leq a,b,c,d\leq 1$. Chứng minh rằng:
$\frac{a}{bcd+2}+\frac{b}{cda+2}+\frac{c}{abd+2}+\frac{d}{abc+2}\leq 1+\frac{1}{abcd+2}$
$\frac{a}{bcd+2}+\frac{b}{cda+2}+\frac{c}{abd+2}+\frac{d}{abc+2}\leq \frac{a+b+c+d}{abcd+2}\leq \frac{ab+1+cd+1}{abcd+2}\leq \frac{abcd+1+2}{abcd+2}=1+\frac{1}{abcd+2} \blacksquare$
$\begin{Bmatrix} 0\leq a,b,c,d\leq 1 & & \\ (a-1)(b-1)\geq 0\Leftrightarrow ab+1\geq a+b & & \\ (c-1)(d-1)\geq 0\Leftrightarrow cd+1\geq c+d & & \\ (ab-1)(cd-1)\geq 0\Leftrightarrow abcd+1\geq ab+cd \end{Bmatrix}$
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