Cần chứng minh: $\sum\limits \dfrac{a^2}{(2a+b)(2a+c)}\le \dfrac{1}{3}$

$\dfrac{a^2}{(2a+b)(2a+c)}=\dfrac{a^2}{4a^2+2ac+2ab+bc}=a^2 . \dfrac{1}{(2a^2+bc)+a(a+b+c)+a(a+b+c)}$

Thật vậy áp dụng BĐT Cauchy SChwars có:

$\dfrac{1}{(2a^2+bc)+a(a+b+c)+a(a+b+c)}\le \dfrac{1}{9}\left(\dfrac{1}{2a^2+bc}+\dfrac{1}{a(a+b+c)}+\dfrac{1}{a(a+b+c)}\right)\\\to \dfrac{a^2}{(2a+b)(2a+c)}\le \dfrac{1}{9}\left(\dfrac{2a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)\\\to \sum\limits \dfrac{a^2}{(2a+b)(2a+c)}\le \dfrac{1}{9}\left(\dfrac{2(a+b+c)}{a+b+c}+\sum\limits \dfrac{a^2}{2a^2+bc}\right)\\\le \dfrac{1}{9}\left(2+\sum\limits \dfrac{a^2}{2a^2+bc}\right)$

Do đó để có đpcm thì chỉ cần chứng minh:

$\sum\limits \dfrac{a^2}{2a^2+bc}\le 1$

Đặt: $A=\sum\limits \dfrac{a^2}{2a^2+bc}$

$\to  2A=\sum\limits \dfrac{2a^2}{2a^2+bc}\\\to 3-2A=\sum\limits \dfrac{bc}{2a^2+bc}\\=\sum\limits \dfrac{b^2c^2}{2a^2bc + b^2c^2}$

Áp dụng BĐT Cauchy có:

$b^2+c^2\ge 2bc\\\to \dfrac{b^2c^2}{2a^2bc+b^2c^2}\ge \dfrac{b^2c^2}{a^2b^2+a^2c^2+b^2c^2}\\\to \sum\limits \dfrac{b^2c^2}{2a^2bc+b^2c^2}\ge \dfrac{a^2b^2+b^2c^2+a^2c^2}{a^2b^2+b^2c^2+a^2c^2}=1\\\to 3-2A\ge 1\\\to A\le 1$

Do vậy hoàn tất chứng minh.

Dấu "=' xảy ra khi: $a=b=c$