Prove this geometric inequality

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 $$frac{-a-b+sqrt{a^2+14ab+b^2}}{2}+a+bgeq3cdotfrac{-a-b+sqrt{a^2+14ab+b^2}}{2}$$ or $$2(a+b)geqsqrt{a^2+14ab+b^2},$$ which is true by AM-GM: $$sqrt{a^2+14ab+b^2}=sqrt{(a+b)^2+12ab}leqsqrt{(a+b)^2+12left(frac{a+b}{2}right)^2}=2(a+b).$$

Let $K$ and $N$ be placed on the sides $AB$ and $AC$ respectively.

Also, let $frac{AK}{AB}>frac{AN}{AC}$ and $Gin NC$ such that $KG||BC$, $KGcap AD={P}$ and $KGcap AE={Q}$.

Thus, $KP=PQ=QG$, which says that it's enough to solve our problem, when $Kequiv B$.

Now, let $AN=kAC$, where $0<k<1$ and $Fin EC$ 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 $Gin NC$ 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 $Iin DE$ such that $MI||AD$.

Thus, $$frac{LM}{BM}=frac{DI}{BI}=frac{DI}{2DI+EI}=frac{frac{DI}{EI}}{frac{2DI}{EI}+1}=$$ $$=frac{frac{AM}{ME}}{2cdotfrac{AM}{ME}+1}=frac{frac{3k}{2(1-k)}}{2cdotfrac{3k}{2(1-k)}+1}=frac{3k}{2(2k+1)}.$$ Id est, $$frac{LM}{BN}=frac{LM}{BM}cdotfrac{BM}{BN}=frac{3k}{2(2k+1)}cdotfrac{2}{2+k}=frac{3k}{(2k+1)(k+2)}$$ and it's enough to prove that: $$frac{3k}{(2k+1)(k+2)}leqfrac{1}{3}$$ or $$(2k+1)(k+2)geq9k,$$ which is true by AM-GM: $$(2k+1)(k+2)geq3sqrt[3]{k^2}cdot3sqrt[3]k=9k.$$