Có $\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}$
$=\dfrac{a^4}{a^2b}+\dfrac{b^4}{b^2c}+\dfrac{c^4}{c^2a}$
Áp dụng bất đẳng thức Schawrtz có:
$\dfrac{a^4}{a^2b}+\dfrac{b^4}{b^2c}+\dfrac{c^4}{c^2a}≥\dfrac{(a^2+b^2+c^2)^2}{a^2b+b^2c+c^2a}$
Ta có: $a^2+b^2+c^2=(a+b+c)(a^2+b^2+c^2)=a^3+ab^2+b^3+bc^2+c^3+ca^2+a^2b+b^2c+c^2a$
$≥2a^2b+2b^2c+2c^2a+a^2b+b^2c+c^2a=3.(a^2b+b^2c+c^2a)$
$⇒a^2b+b^2c+c^2a≤\dfrac{a^2+b^2+c^2}{3}$
$⇒\dfrac{a^4}{a^2b}+\dfrac{b^4}{b^2c}+\dfrac{c^4}{c^2a}≥\dfrac{(a^2+b^2+c^2)^2}{\dfrac{a^2+b^2+c^2}{3}}=3.(a^2+b^2+c^2)$
$⇒\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}≥3.(a^2+b^2+c^2)(đpcm)$
Dấu $=$ xảy ra $⇔a=b=c=\dfrac{1}{3}$
