Cho $a_{1}, a_{2},...,a_{n}\epsilon [0,1]$.
Chứng minh:$\frac{a_1}{a_{2}+a_3+...a_n+1}+\frac{a_2}{a_1+a_3+...+a_n+1}+...+\frac{a_n}{a_1+a_2+...+a_{n-1}} + (1-a_1)(1-a_2)...(1-a_n)\leq 1$.
Cho $a_{1}, a_{2},...,a_{n}\epsilon [0,1]$.
Chứng minh:$\frac{a_1}{a_{2}+a_3+...a_n+1}+\frac{a_2}{a_1+a_3+...+a_n+1}+...+\frac{a_n}{a_1+a_2+...+a_{n-1}} + (1-a_1)(1-a_2)...(1-a_n)\leq 1$.
Cho $a_{1}, a_{2},...,a_{n}\epsilon [0,1]$.
Chứng minh:$\frac{a_1}{a_{2}+a_3+...a_n+1}+\frac{a_2}{a_1+a_3+...+a_n+1}+...+\frac{a_n}{a_1+a_2+...+a_{n-1}} + (1-a_1)(1-a_2)...(1-a_n)\leq 1$.
Không mất tính tổng quát, giả sử $a_{1}\geq a_{2}\geq ...\geq a_{n}$. Khi đó, ta có:
$VT\leq \frac{a_{1}}{a_{2}+a_{3}+...+a_{n}+1}+\frac{a_{2}}{a_{2}+a_{3}+...+a_{n}+1}+...+\frac{a_{n}}{a_{2}+a_{3}+...+a_{n}+1}=\frac{a_{1}+...+a_{n}}{1+a_{2}+...+a_{n}}=1-\frac{1-a_{1}}{1+a_{2}+...+a_{n}}$
Như vậy, ta cần chứng minh:
$(1-a_1)(1-a_2)...(1-a_n)-\frac{1-a_{1}}{a_{2}+a_{3}+...+a_{n}+1}\leq 0$
$\Leftrightarrow (1-a_2)...(1-a_n)(1+a_{2}+a_{3}+...+a_{n})\leq 1$ ( luôn đúng theo AM-GM)
$\Rightarrow Q.E.D$
Edited by cristianoronaldo, 24-07-2017 - 16:54.
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