<s>[INST] 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.
 [/INST] 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:
$$\sqrt{a^2+14ab+b^2}=\sqrt{(a+b)^2+12ab}\leq\sqrt{(a+b)^2+12\left(\frac{a+b}{2}\right)^2}=2(a+b).$$
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