Let a,b,c be positive real number, proof.
Solution 1:
since $$(x+y)^2\ge 4xy$$ so $$(a+b+2c)^2=(a+c+b+c)^2\ge 4(a+c)(b+c)$$ so $$\sum_{cyc}\dfrac{16}{(a+b+2c)^2}\le\sum_{cyc}\dfrac{4}{(a+c)(b+c)}=\dfrac{8(a+b+c)}{(a+b)(b+c)(a+c)}$$ and use this well know inequality $$(a+b)(b+c)(c+a)\ge\dfrac{8}{9}(a+b+c)(ab+bc+ac)$$ so $$\dfrac{8(a+b+c)}{(a+b)(b+c)(a+c)}\le\dfrac{9}{ab+bc+ac}\cdots (1)$$ since $$a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Longrightarrow 3(ab+bc+ac)=3abc(a+b+c)\le (ab+bc+ac)^2$$ so $$ab+bc+ac\ge 3$$ so $$\sum_{cyc}\dfrac{16}{(a+b+2c)^2}\le\dfrac{9}{ab+bc+ac}\le 3$$