Giải thích các bước giải:
1.Ta có:
$(a-b)^2+(b-c)^2+(c-a)^2\ge0$
$\to a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge 0$
$\to 2(a^2+b^2+c^2-ab-bc-ca)\ge 0$
$\to a^2+b^2+c^2-ab-bc-ca\ge 0$
$\to (a^2-ab)+(b^2-bc)+(c^2-ca)\ge 0$
$\to a(a-b)+b(b-c)+c(c-a)\ge 0$
2.Ta có:
$(a-b)^2\ge0$
$\to a^2-2ab+b^2\ge 0$
$\to a^2+b^2\ge 2ab$
$\to 2(a^2+b^2)\ge a^2+b^2+2ab$
$\to 2(a^2+b^2)\ge(a+b)^2$
$\to \dfrac{a^2+b^2}{2}\ge (\dfrac{a+b}{2})^2$
3.Ta có:
$\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{9}{(x+y)+(y+z)+(z+x)}=\dfrac9{2(x+y+z)}$
$\to (x+y+z)(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x})\ge\dfrac92$
