Prove this geometric inequality Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$
My attempt: I think cross ratio can be used to prove it though I am not sure how.
Let $a=KL,b=LM,c=MN$
We have to proove $a+b+c\geq 3z$, i.e. $a+b\geq 2c$
Since $R(K,N;L,M)=R(B,C;D,E)$ so $R(K,N;L,M)= \frac{1}{2}:\frac{1}{2}$ thus $4ac=(a+b)(b+c)$.
I am not able to move further from here.

Let $K$ and $N$ be placed on the sides $AB$ and $AC$ respectively.
Also, let $\frac{AK}{AB}>\frac{AN}{AC}$ and $G\in NC$ such that $KG||BC$, $KG\cap AD=\{P\}$ and $KG\cap AE=\{Q\}$.
Thus, $KP=PQ=QG$, which says that it's enough to solve our problem, when $K\equiv B$.
Now, let $AN=kAC$, where $0<k<1$ and $F\in EC$ such that $NF||AE.$
Thus, $$\frac{BM}{BN}=\frac{BE}{BF}=\frac{\frac{2}{3}BC}{\frac{2}{3}BC+\frac{k}{3}BC}=\frac{2}{2+k}.$$
Now, let $G\in NC$ such that $EG||BN.$
Thus, $$\frac{AM}{ME}=\frac{AN}{NG}=\frac{kAC}{\frac{2}{3}(1-k)AC}=\frac{3k}{2(1-k)}.$$
Now, let $I\in DE$ such that $MI||AD$.
Thus, $$\frac{LM}{BM}=\frac{DI}{BI}=\frac{DI}{2DI+EI}=\frac{\frac{DI}{EI}}{\frac{2DI}{EI}+1}=$$
$$=\frac{\frac{AM}{ME}}{2\cdot\frac{AM}{ME}+1}=\frac{\frac{3k}{2(1-k)}}{2\cdot\frac{3k}{2(1-k)}+1}=\frac{3k}{2(2k+1)}.$$
Id est, $$\frac{LM}{BN}=\frac{LM}{BM}\cdot\frac{BM}{BN}=\frac{3k}{2(2k+1)}\cdot\frac{2}{2+k}=\frac{3k}{(2k+1)(k+2)}$$ and it's enough to prove that:
$$\frac{3k}{(2k+1)(k+2)}\leq\frac{1}{3}$$ or $$(2k+1)(k+2)\geq9k,$$ which is true by AM-GM:
$$(2k+1)(k+2)\geq3\sqrt[3]{k^2}\cdot3\sqrt[3]k=9k.$$
