Giải thích các bước giải:

Theo BĐT Cô - si ta có :

$\sqrt[]{\dfrac{2c}{a+b}} = \dfrac{4c}{2.\sqrt[]{2c.(a+b)}} ≥ \dfrac{4c}{2c+a+b}$

Theo BĐT Svac - sơ lại có :

$\bigg(\dfrac{a}{b+c} + 1\bigg) + \bigg(\dfrac{b}{c+a} + 1\bigg) - 2$

$ = \dfrac{a+b+c}{b+c} + \dfrac{a+b+c}{c+a} - 2$

$ = (a+b+c).\bigg(\dfrac{1}{b+c} + \dfrac{1}{c+a}\bigg) - 2$

$≥ (a+b+c).\dfrac{4}{2c+a+b} - 2$

$\dfrac{4.(a+b+c)}{2c+a+b} - 2$

Khi đó ta có : $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \sqrt[]{\dfrac{2c}{a+b}} $

$≥\dfrac{4.(a+b+c)}{2c+a+b} + \dfrac{4c}{2c+a+b} - 2$

$ = \dfrac{4.(a+b+2c)}{a+b+2c} - 2$

$ = 4-2=2$

$⇒ T ≥ 2021. 2 = 4042$

Dấu "=" xảy ra khi $a=b=c$

Vậy $Min T = 4042$ khi $a=b=c$

`\sqrt((2c)/(a+b))≥\sqrt((2c×2c)/(2c×(a+b)))≥(2c)/(\sqrt(2c×(a+b)))≥(4c)/(2\sqrt(2c×(a+b)))≥(4c)/(2c+b+a)`

`(a/(b+c)+b/(c+a)+\sqrt((2c)/(a+b))≥(a+b+c)/(b+c)+(a+b+c)/(c+a)+(4c)/(2c+b+a)-2`

`⇔T/(2021)≥(a+b+c)/((b+c)+(c+a))+(4c)/(2c+b+a)-2`

`⇔T/(2021)≥(4(a+b+c))/(a+b+2c)+4c/(2c+b+a)-2`

`⇔T/(2021)≥(4(a+b+2c)/(a+b+2c)-2`

`⇔T/(2021)≥4-2`

`⇔T/(2021)≥2`

`⇔T≥4042`

`''=''`xẩy ra khi :
`a=b=c`