It takes me a lot a time to solve this :
Let $a,b,c>0$ with $a\geq b\geq c$ then we have : $$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$$
My proof : We need some lemmas to begin :
First lemma : Let $x,y>0$ then we have : $$\frac{x}{x+\frac{1}{2}}+\frac{y}{y+\frac{1}{2}}=\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}+\frac{p}{p+\frac{1}{2}}$$ Where $p$ is equal to: $$\frac{\frac{x}{(x+\frac{1}{2})^2}+\frac{y}{(y+\frac{1}{2})^2}}{\frac{1}{(x+\frac{1}{2})^2}+\frac{1}{(y+\frac{1}{2})^2}}$$
the proof is easy it's just calculus .
Second lemma : With the notation of the first lemma we have : $$\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}\geq \frac{p}{p+\frac{1}{2}}$$
Proof :
With the first lemma and the concavity of $\frac{x}{x+\frac{1}{2}}$ and the use of Jensen's inequality :
$$\frac{x}{x+\frac{1}{2}}+\frac{y}{y+\frac{1}{2}}-\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}=\frac{p}{p+\frac{1}{2}}\leq \frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}} $$
Now we attack the problem :
We prove a stronger result :
We have with $x=\frac{1}{a}$ and $y=\frac{1}{b}$ and $z=\frac{1}{c}$:
$$\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a}=\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x}$$
But :
$$\frac{xy}{2x+y}+\frac{yz}{2z+y}=y(\frac{\frac{1}{2}\frac{x}{y}}{\frac{x}{y}+\frac{1}{2}}+\frac{\frac{1}{2}\frac{z}{y}}{\frac{z}{y}+\frac{1}{2}})$$
Now we apply the first lemma to get :
$$y(\frac{\frac{1}{2}\frac{x}{y}}{\frac{x}{y}+\frac{1}{2}}+\frac{\frac{1}{2}\frac{z}{y}}{\frac{z}{y}+\frac{1}{2}})=\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}}+Q$$
Finally we have :
$$\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x}=\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}}+Q+R+T$$ We multiply by $\frac{x+y+z}{xy+yz+zx}$ : $$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x})=\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})+\frac{x+y+z}{xy+yz+zx}(Q+R+T)$$
But :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})=1$$
And with the second lemma we have :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})=1\geq\frac{x+y+z}{xy+yz+zx}(Q+R+T)$$
So we prove :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x})\leq 2$$
Or $$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a})\leq 2$$
But :
$$\sum_{cyc}\frac{1}{2a+b}-\sum_{cyc}\frac{1}{2b+a}= -\frac{(2 (a - b) (a - c) (b - c) (2 a^2 + 7 a b + 7 a c + 2 b^2 + 7 b c + 2 c^2))}{((2 a + b) (a + 2 b) (2 a + c) (a + 2 c) (2 b + c) (b + 2 c))}$$
And with the condition $a\geq b\geq c$ it's negative so :
$$2\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a})\leq 2$$
Done !
My question is have you an alternative proof ?
Thanks a lot for your time !
As an alternative, while not shorter, it can be brute-forced with just elementary algebra . . .
Since the $\text{LHS}$ is homogeneous, we can assume $c=1$, and $a\ge b \ge 1$.
Replacing $c$ by $1$, and simplifying, we get $$ 1-\text{LHS} = \frac {-a^2b+2a^3b-a^2-ab^2-a^2b^2-ab+2b^3-b^2+2a} {(a+b+1)(2a+b)(2b+1)(2+a)} $$ so it remains to show $$ -a^2b+2a^3b-a^2-ab^2-a^2b^2-ab+2b^3-b^2+2a\ge 0 $$ Replacing $b$ by $1+x$ and $a$ by $1+x+y$, the $\text{LHS}$ of the above inequality simplifies to $$ 3x^2+3xy+3y^2+9x^2y+9y^2x+4x^3y+5y^2x^2+2y^3x+4x^3+2y^3+x^4 $$ which is nonnegative since all coefficients are positive and $x,y\ge 0$.