$\text{Ta chứng minh }$
$\text{3.(a²+b²+c²)≥(a+b+c)² (1) }$
$\text{⇔3.(a²+b²+c²)≥a²+b²+c²+2ab+2bc+2ca }$
$\text{⇔2(a²+b²+c²)≥2ab+2bc+2ca }$
$\text{⇔(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)≥0 }$
$\text{⇔(a-b)²+(b-c)²+(c-a)²≥0 (luôn đúng) }$
$\text{⇒DPCM }$
$\text{Ta chứng minh }$
$\text{(a+b+c)²≥3(ab+bc+ca) (2) }$
$\text{⇔a²+b²+c²+2ab+2bc+2ca≥3(ab+bc+ca) }$
$\text{⇔a²+b²+c²≥ab+bc+ca }$
$\text{⇔2.(a²+b²+c²)≥2.(ab+bc+ca) }$
$\text{⇔(a-b)²+(b-c)²+(c-a)²≥0 }$
$\text{Luôn đúng⇒đpcm }$
$\text{(1);(2)⇒3.(a²+b²+c²)≥(a+b+c)²≥3(ab+bc+ca) (dpcm) }$
