Với $a,b,c>0$ t/m: $a+b+c+ab+bc+ca=6abc$.Chứng minh:
$P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq3$.
Từ giả thiết a+b+c+ab+bc+ca=6abc thì bạn chia tất cả cho abc .
sau đó áp dụng bđt x^2 +y^2+z^2 $\geq $xy+yz+zx    và x^2+1 $\geq$ 2x
$\frac{1}{a^{2}}$ + $\frac{1}{b^{2}}$ +$\frac{1}{c^{2}}$ $\geq$ $\frac{1}{ab}$ + $\frac{1}{bc}$ + $\frac{1}{ca}$ 
1a2 +1 + 1b2 +1+1c2 +1 $\geq $ 2( $\frac{1}{b}$ +$\frac{1}{a}$+$\frac{1}{c}$) 
=> 2( 1/a^2 +1/b^2+1/c^2) + (1/a^2 +1/b^2+1/c^2 +3) $\geq$ 2( $\frac{1}{b}$+ $\frac{1}{a}$ +$\frac{1}{c}$+ $\frac{1}{ab}$ +$\frac{1}{bc}$+ $\frac{1}{ca}$)
=>sau đó tự làm nốt =>dpcm
dấu "=" xảy ra khi a=b=c=1 –  lovesomebody121 22-02-16 09:05 PM
$\frac 32P+\frac 32=\frac 12[(\frac{1}{a^2}+1)+(\frac{1}{b^2}+1)+(\frac{1}{c^2}+1)]+(\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2})$
$ \ge \frac 1a+\frac 1b+\frac 1c+\frac 1{ab} + \frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c+ab+bc+ca}{abc}=6$
$\Leftrightarrow 3P+3 \ge 12\Leftrightarrow P \ge 3$

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