Question

cho abc = 1 và \(a^3>36\). Cmr: \(\frac{a^2}{3}+b^2+c^2\ge ab+bc+ca\)

Answer 1

\(\Leftrightarrow\frac{a^2}{3}+b^2+c^2+2bc-3bc-a\left(b+c\right)\ge0\)

\(\Leftrightarrow\frac{a^2}{4}+\left(b+c\right)^2-a\left(b+c\right)+\frac{a^2}{12}-3bc\ge0\)

\(\Leftrightarrow\left(\frac{a}{2}-b-c\right)^2+\frac{a^3-36abc}{12a}\ge0\)

\(\Leftrightarrow\left(\frac{a}{2}-b-c\right)^2+\frac{a^3-36}{12a}\ge0\)

Mà \(a^3>36\Rightarrow\left\{{}\begin{matrix}a>0\\a^3-36>0\end{matrix}\right.\)

\(\Rightarrow\left(\frac{a}{2}-b-c\right)^2+\frac{a^3-36}{12a}>0\)

Dấu "=" ko xảy ra nên BĐT đã cho sai