cho `a;b;c` là `3` số dương thõa mãn` :1/(a+1)+1/(b+1)+1/(c+1)=2` tìm `maxQ=abc`

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1. `1/(a+1)+1/(b+1)+1/(c+1)≥(9)/(1+1+1+a+b+c)`

   `⇔2≥(9)/(3+a+b+c)`

   `⇔6+2a+2b+2c≥9`

   `⇔a+b+c≥3/2`

   ⇔\begin{cases} \frac{1}{a+1}+ \frac{1}{b+1}+ \frac{1}{c+1}=2\\\frac{1}{b+1}+\frac{1}{a+1}+ \frac{1}{c+1}=2\\\frac{1}{c+1}+ \frac{1}{b+1}+ \frac{1}{c+1}=2 \end{cases}

   ⇔\begin{cases} \frac{1}{a+1}=(1- \frac{1}{b+1})+(1- \frac{1}{c+1})\\\frac{1}{b+1}=(1- \frac{1}{a+1})+(1- \frac{1}{c+1})\\\frac{1}{c+1}=(1- \frac{1}{b+1})+(1- \frac{1}{a+1}) \end{cases}

   ⇔\begin{cases} \frac{1}{a+1}=\frac{b}{b+1}+\frac{c}{c+1}\\\frac{1}{b+1}=\frac{a}{a+1}+ \frac{c}{c+1}\\\frac{1}{c+1}=\frac{b}{b+1}+ \frac{a}{a+1} \end{cases}

   ⇔\begin{cases} \frac{1}{a+1}≥\sqrt(\frac{bc}{(b+1)(c+1)})\\\frac{1}{b+1}≥\sqrt(\frac{ac}{(a+1)(c+1)})\\\frac{1}{c+1}≥ \sqrt(\frac{ba}{(b+1)(a+1)})\end{cases}

   `⇒1/((a+1)(b+1)(c+1))≥8\sqrt((a^2b^2c^2)/((a+b)(b+c)(c+a))^2)`

   `⇒1/((a+1)(b+1)(c+1))≥8(abc)/((a+b)(b+c)(c+a))`

   `⇒1≥8abc`

   `⇒1/8 ≥abc`

   `”=”` xẩy ra khi :

   \begin{cases}a+b+c=3/2 \\a=b=c \end{cases}

   `⇒a=b=c=1/2`

   vậy `maxQ=1/8` khi `a=b=c=1/2`

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