Ta có:\(\frac{a^2}{3}+b^2+c^2>ab+bc+ca\)
\(\Leftrightarrow\) \(\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0\)
\(\Leftrightarrow\) \(\frac{a^2}{4}+\frac{a^2}{12}+b^2+c^2-ab-ca+2bc-3bc>0\)
\(\Leftrightarrow\) \(\left(\frac{a^2}{4}+b^2+c^2-ab-ca+2bc\right)+\frac{a^2}{12}-3bc>0\)
\(\Leftrightarrow\) \(\left(\frac{a}{2}-b-c\right)^2+\frac{a^2}{12}-3bc>0\)
\(\Leftrightarrow\) \(\left(\frac{a}{2}-b-c\right)^2+\frac{a^3-36abc}{12a}>0\)
Vì : abc=1 và \(a^3>36\)
\(\Rightarrow\frac{a^3-36abc}{12a}>0\)
Mà:\(\left(\frac{a}{2}-b-c\right)^2\ge0\forall a;b;c\)
\(\Rightarrow\left(\frac{a}{2}-b-c\right)^2+\frac{a^3-35abc}{12a}>0\)
Hay: \(\frac{a^2}{3}+b^2+c^2>ab+bc+ca\)(đpcm)
