áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)
Cần chứng minh rằng \(\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
áp dụng bđt Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\) ( đpcm )
Vậy \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge a^2+b^2+c^2\)( đpcm )
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}=\dfrac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{ab+bc+ca}\ge a^2+b^2+c^2\)
