a) ⇔ $a^2b + ab^2 - 2abc + b^2c + bc^2 - 2abc + a^2c + ac^2 - 2abc \geq 0 $
⇔ $(a^2b + bc^2 - 2abc) + (ab^2 +ac^2 - 2abc) + (b^2c + a^2c - 2abc) \geq 0$
⇔ $b(a^2 - 2ac + c^2) + a(b^2 - 2bc + c^2) + c(b^2 - 2ab + a^2) \geq 0$
⇔ $b(a - c)^2 + a(b - c)^2 + c(a - b)^2 \geq 0$ (luôn đúng do a, b, c ≥ 0)
Dấu "=" xảy ra ⇔ $a = b = c$
b) ⇔ $\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab} \geq 2\bigg(\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c}\bigg)$
⇔ $\dfrac{a^2 + b^2 + c^2}{abc} - 2\dfrac{bc + ac - ab}{abc}\geq 0$
⇔ $\dfrac{a^2 + b^2 + c^2 + 2ab - 2bc - 2ac}{abc} \geq 0$
⇔ $\dfrac{(a + b - c )^2}{abc} \geq 0$ (luôn đúng do a,b,c > 0)
c) ⇔ $\dfrac{a^2c + ab^2 + bc^2}{abc} \geq \dfrac{b^2c + ac^2 + a^2b}{abc}$
⇒$ a^2c + ab^2 + bc^2 \geq b^2c + ac^2 + a^2b$ (Vì $abc > 0$ do a,b,c >0)
⇔ $a^2c + ab^2 + bc^2 - b^2c - ac^2 - a^2b \geq 0$
⇔ $(c-b)(b-a)(c-a) \geq 0$ (Luôn đúng do a ≤ b ≤c)
