\(4b^2c^2-\left(b^2+c^2-a^2\right)\)
\(\Leftrightarrow\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(\Leftrightarrow\left[\left(b+c\right)^2-a^2\right]\left[a^2-\left(b-c\right)^2\right]\)
\(\Leftrightarrow\left(a+b+c\right)\left(b+c-a\right)\left(a+b-c\right)\left(a+c-b\right)\)
Ta có: a,b,c là 3 cạnh của 1 tam giác
\(\Rightarrow\cdot a+b+c>0\)
\(\cdot a+b>c\Rightarrow a+b-c>0\)
\(\cdot b+c>a\Rightarrow b+c-a>0\)
\(\cdot a+c>b\Rightarrow a+c-b>0\)
\(\Rightarrow\left(a+b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)>0\)
Vậy \(4b^2c^2-\left(b^2+c^2-a^2\right)>0\)
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