Đáp án: $P\ge \dfrac{2021}{3}$
Giải thích các bước giải:
Ta có:
$P=\dfrac{1}{a^2+b^2+c^2}+\dfrac{2020}{ab+bc+ca}$
$\to P=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{2018}{ab+bc+ca}$
$\to P\ge \dfrac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\dfrac{2018}{\dfrac13(a+b+c)^2}$
$\to P\ge \dfrac{9}{(a+b+c)^2}+\dfrac{2018\cdot 3}{(a+b+c)^2}$
$\to P\ge \dfrac{9}{3^2}+\dfrac{2018\cdot 3}{3^2}$
$\to P\ge \dfrac{2021}{3}$
Dấu = xảy ra khi $a=b=c=1$
