Giải thích các bước giải:

Trước hết ta chứng minh bổ đề :

$\sqrt{a+x^2}+\sqrt{b+y^2}\le\sqrt{2(a+b)+(x+y)^2}\leftrightarrow 4(x-y)(bx-ay)\le (a-b)^2$

Thật vậy ta có :
$\sqrt{a+x^2}+\sqrt{b+y^2}\le\sqrt{2(a+b)+(x+y)^2}$

$\leftrightarrow (\sqrt{a+x^2}+\sqrt{b+y^2})^2\le(\sqrt{2(a+b)+(x+y)^2})^2$

$\leftrightarrow a+x^2+b+y^2+2\sqrt{a+x^2}.\sqrt{b+y^2}\le2(a+b)+(x+y)^2$

$\leftrightarrow 2\sqrt{a+x^2}.\sqrt{b+y^2}\le a+b+2xy$

$\leftrightarrow (2\sqrt{a+x^2}.\sqrt{b+y^2})^2\le (a+b+2xy)^2$

$\leftrightarrow 4(a+x^2)(b+y^2)\le 4axy+4bxy+4x^2y^2+a^2+2ab+b^2$

$\leftrightarrow 4ab+4ay^2+4bx^2+4x^2y^2\le 4axy+4bxy+4x^2y^2+a^2+2ab+b^2$

$\leftrightarrow 4(x-y)(bx-ay)\le (a-b)^2$

$\to đpcm$

Ta có :
$T=\sqrt{12a+(b-c)^2}+\sqrt{12b+(c-a)^2}$

$=\sqrt{a+(\dfrac{b-c}{\sqrt{12}})^2}+\sqrt{b+(\dfrac{c-a}{\sqrt{12}})^2}$

Lại có :
$4(\dfrac{b-c}{\sqrt{12}}-\dfrac{a-c}{\sqrt{12}})(b.\dfrac{b-c}{\sqrt{12}}-a.\dfrac{a-c}{\sqrt{12}})$

$=4\dfrac{b-a}{\sqrt{12}}.\dfrac{b^2-cb-a^2+ac}{\sqrt{12}}$

$=4\dfrac{b-a}{\sqrt{12}}.\dfrac{(b^2-a^2)-(cb-ac)}{\sqrt{12}}$

$=4\dfrac{b-a}{\sqrt{12}}.\dfrac{(b+a)(b-a)-c(b-a)}{\sqrt{12}}$

$=4\dfrac{b-a}{\sqrt{12}}.\dfrac{(b+a-c)(b-a)}{\sqrt{12}}$

$=\dfrac{(b+a-c)(b-a)^2}{3}$

$=\dfrac{(3-2c)(b-a)^2}{3}\le (b-a)^2$ Vì $a,b,c\ge 0$

$\to T\le \sqrt{2(a+b)+(\dfrac{b-c+a-c}{\sqrt{12}})^2}$

$\to T\le \sqrt{2(3-c)+\dfrac{(3-c-2c)^2}{12}}$

$\to T\le \sqrt{\dfrac{12(2(3-c))+(3-3c)^2}{12}}$

$\to T\le \sqrt{\dfrac{9c^2-42c+81}{12}}$

$\to T\le \sqrt{\dfrac{9c^2-30c+25}{12}}$

$\to T\le \sqrt{\dfrac{(5-3c)^2}{12}}$

$\to T\le \dfrac{5-3c}{\sqrt{12}}$

$\to \sqrt{a+(\dfrac{b-c}{\sqrt{12}})^2}+\sqrt{b+(\dfrac{c-a}{\sqrt{12}})^2}\le \dfrac{5-3c}{\sqrt{12}}$

Tương tự :
$\sqrt{b+(\dfrac{a-c}{\sqrt{12}})^2}+\sqrt{c+(\dfrac{a-b}{\sqrt{12}})^2}\le \dfrac{5-3a}{\sqrt{12}}$

$\sqrt{a+(\dfrac{b-c}{\sqrt{12}})^2}+\sqrt{c+(\dfrac{a-b}{\sqrt{12}})^2}\le \dfrac{5-3b}{\sqrt{12}}$

Cộng vế với vế

$\to 2(\sqrt{a+(\dfrac{b-c}{\sqrt{12}})^2}+\sqrt{b+(\dfrac{c-a}{\sqrt{12}})^2}+\sqrt{c+(\dfrac{a-b}{\sqrt{12}})^2})\le \dfrac{15-3(a+b+c)}{\sqrt{12}}$

$\to \sqrt{12a+(b-c)^2}+\sqrt{12b+(c-a)^2}+\sqrt{12c+(a-b)^2}\le 3$

Dấu = xảy ra khi $a=b=0,c=3$