Xét:

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

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

$= \dfrac{4a^3+2a^2b+2a^2c+abc - 3a^3-3a^2b-3a^2c}{(a+b+c)(2a+b)(2a+c)}$

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

Ta có:

$\sum_{cyc} \dfrac{a}{a+b+c} - 3\sum_{cyc} \dfrac{a^2}{(2a+b)(2a+c)} = \dfrac{1}{a+b+c} \sum_{cyc} \dfrac{a(a-b)(a-c)}{(2a+b)(2a+c)}$

$\Leftrightarrow 1 - 3\sum_{cyc} \dfrac{a^2}{(2a+b)(2a+c)} = \dfrac{1}{a+b+c} \sum_{cyc} \dfrac{a(a-b)(a-c)}{(2a+b)(2a+c)}$

Không mất tính tổng quát, giả sử $a \ge b \ge c > 0$.

Ta có $c-a \le 0$, $c-b \le 0 \Rightarrow \dfrac{c(c-a)(c-b)}{(2c+a)(2c+b)} \ge 0$.

Xét tổng 2 số hạng còn lại:

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

$ = (a-b) \left[ \dfrac{a(a-c)}{(2a+b)(2a+c)} - \dfrac{b(b-c)}{(2b+a)(2b+c)} \right]$

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

$ = (a-b). \dfrac{2ab(a^2-b^2) + c(a^3-b^3) + 4abc(a-b) - c^2(a^2-b^2)}{(2a+b)(2a+c)(2b+a)(2b+c)}$

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

Do $a \ge b \ge c > 0$ nên $2ab \ge 2c^2 > c^2 \Rightarrow 2ab(a+b) - c^2(a+b) > 0$. $\Rightarrow S \ge 0$.

Cộng 3 số hạng lại ta được:

$\sum_{cyc} \dfrac{a(a-b)(a-c)}{(2a+b)(2a+c)} \ge 0$

$\Rightarrow 1 - 3\sum_{cyc} \dfrac{a^2}{(2a+b)(2a+c)} \ge 0$

$\Leftrightarrow \sum_{cyc} \dfrac{a^2}{(2a+b)(2a+c)} \le \dfrac{1}{3}$

Dấu "=" xảy ra $\Leftrightarrow a=b=c$.

Vậy đpcm.