Chứng minh bất đẳng thức (Trích đề thi HSG)

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viethoang1999

$b+c>a$
\Rightarrow $a+b+c\ge 2a$
\Rightarrow $a<1$
Tương tự \Rightarrow $(1-a)(1-b)(1-c)>0$
\Leftrightarrow $1-\left ( a+b+c \right )+ab+bc+ac-abc> 0$
\Rightarrow $2\left ( ab+bc+ac \right )> 2+abc$
\Rightarrow $2\left ( ab+bc+ac \right )+a^2+b^2+c^2> a^2+b^2+c^2+2abc+2$
 
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