$\text{Ta sẽ chứng minh :2.(a²-ab+b²)²≥ $a^{4}$ +$b^{4}$ }$
$\text{ ⇔2[(a²+b²)-ab]²≥ $a^{4}$ +$b^{4}$ }$
$\text{ ⇔2.[ (a²+b²)²-2.ab(a²+b²)+a².b²]≥ $a^{4}$ +$b^{4}$ }$
⇔2.(a⁴+2.a².b²+b⁴-2ab(a²+b²)+a²b² )≥$a^{4}$ +$b^{4}$
$\text{⇔2.a⁴+4.a²b²+2.b⁴-4.ab(a²+b²)+2a²b²≥$a^{4}$ +$b^{4}$ }$
$\text{⇔a⁴+4.a²b²+b⁴-4.ab(a²+b²)+2a²b²≥0}$
$\text{⇔(a⁴+2a²b²+b⁴)-4.ab(a²+b²)+(2ab)²≥0}$
$\text{⇔(a²+b²)²-4.ab(a²+b²)+(2ab)²≥0}$
$\text{⇔(a²+b²-2ab)²≥0 (luôn đúng)}$
$\text{Nên 2.(a²-ab+b²)²≥$a^{4}$ +$b^{4}$ (1)}$
$\text{Chứng minh tương tự}$
$\text{2.(b²-bc+c²)²≥$b^{4}$ +$c^{4}$ (2)}$
$\text{2.(c²-ac+a²)²≥$c^{4}$ +$a^{4}$ (3)}$
$\text{Nhân (1);(2);(3) theo vế ta có}$
$\text{8.(a²-ab+b²)².(b²-bc+c²)²(c²-ac+a²)²≥ ($a^{4}$ +$b^{4}$).($b^{4}$ +$c^{4}$).($c^{4}$ +$a^{4}$)=8}$
$\text{⇔(a²-ab+b²).(b²-bc+c²)(c²-ac+a²)≥1 }$
