Với $a,b,c>0$ ta có đánh giá trội sau:

$$\sum_{\rm cyc}\dfrac{a+2b+2c}{\sqrt{a^2+16bc}}>\sum_{\rm cyc}\dfrac{a+2b+2c}{\sqrt{a^2+4ac+4bc+16bc}}=A$$

Theo BĐT AM-GM ta có:

$$A=\sum_{\rm cyc}\dfrac{a+2b+2c}{\sqrt{a+4b)(a+4c)}}\geqslant \sum_{\rm cyc}\dfrac{a+2b+2c}{\dfrac{1}{2}(a+4b+a+4c)}=3$$

Kết thúc chứng minh.