\(\Leftrightarrow\frac{a^2}{3}+b^2+4c^2-ab-2bc-2ca>0\)
\(\Leftrightarrow\frac{a^2}{4}+\left(b^2+4bc+4c^2\right)-a\left(b+2c\right)+\frac{a^2}{12}-6bc>0\)
\(\Leftrightarrow\frac{a^2}{4}+\left(b+2c\right)^2-a\left(b+2c\right)+\frac{a^2-36bc}{12}>0\)
\(\Leftrightarrow\left(\frac{a}{2}-b-2c\right)^2+\frac{a^3-36abc}{12a}>0\)
\(\Leftrightarrow\left(\frac{a}{2}-b-2c\right)^2+\frac{a^3-36}{12a}>0\) (1)
Do \(a^3>36\Rightarrow\left\{{}\begin{matrix}a>0\\a^3-36>0\end{matrix}\right.\) \(\Rightarrow\frac{a^3-36}{12a}>0\)
\(\Rightarrow\left(1\right)\) luôn đúng
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