Prove this geometric inequality

Question

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.

Answer 1

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: $$(2k+1)(k+2)\geq3\sqrt[3]{k^2}\cdot3\sqrt[3]k=9k.$$

Answer 2

Another way.

Let $KL=a$, $LM=x$ and $MN=b$.

Thus, since $$\frac{KM}{KN}:\frac{LM}{LN}=\frac{BE}{BC}:\frac{DE}{DC},$$ we obtain: $$\frac{(a+x)(b+x)}{(a+b+x)x}=\frac{4}{3}$$ or $$x^2+(a+b)x-3ab=0$$ or $$x=\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}$$ and we need to prove that $$\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}+a+b\geq3\cdot\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}$$ or $$2(a+b)\geq\sqrt{a^2+14ab+b^2},$$ which is true by AM-GM: $$\sqrt{a^2+14ab+b^2}=\sqrt{(a+b)^2+12ab}\leq\sqrt{(a+b)^2+12\left(\frac{a+b}{2}\right)^2}=2(a+b).$$