It seems easy. Let $R$ be the circumadius of triangle $LMN$. Of course the point $H$ is it's circumcenter ( a well known fact). Also segments $AL$,$BM$,$CN$ are equal to the sides of triangle $LMN$ thus all we want to do, is to prove the following \[ LM^2+MN^2+NL^2>3R^2 \] and this is easy to prove using using formulas concerning triangles. I think that even the stronger on holds \[ LM^2+MN^2+NL^2>8R^2 \]

Well, that well known fact is wrong ... the orthocenter isn't the center of the Feuerbach circle. The well known fact is that the orthocenter of $LMN$ is the circumcenter of $ABC$ ...

Lemma. $\boxed {\ \mathrm {\ acute}\ \triangle ABC\Longrightarrow 8R^2<a^2+b^2+c^2\le 9R^2\ }\ .$ Proof. $a^2+b^2+c^2-8R^2=4R^2\cdot(\sin ^2A+\sin ^2B+\sin^2C-2)=$ $2R^2[(1-\cos 2A)+(1-\cos 2B)+(1-\cos 2C)-4]=$ $-2R^2(1+\cos 2A+\cos 2B+\cos 2C)=-2R^2[2\cos^2A+2\cos(B+C)\cos(B-C)]=$ $-4R^2[\cos^2A-\cos A\cos (B-C)]=4R^2\cos A[\cos (B+C)+\cos (B-C)]=$ $8R^2\cos A\cos B\cos C>0\ .$ Remark. $a^2+b^2+c^2=8R^2(1+\cos A\cos B\cos C)\le 9R^2$ because it is well-known that in any triangle $ABC$, $\cos A\cos B\cos C\le \frac 18\ .$ The proof of the proposed problem. It is well-knownn that $AH=2R\cos A$ a.s.o. Therefore, $4\cdot HM^2=2\cdot(HB^2+HC^2)-a^2=8R^2(\cos^2B+\cos^2C)-a^2=$ $8R^2(2-\sin^2B+\sin^2C)-a^2=16R^2-2(b^2+c^2)-a^2\ .$ Thus, $4\cdot\sum HM^2=48R^2-5(a^2+b^2+c^2)\ .$ From the above lemma results the relation $\frac 13 (a^2+b^2+c^2)\le 4\cdot\sum HM^2<6(a^2+b^2+c^2)-5(a^2+b^2+c^2)=a^2+b^2+c^2$, i.e. $\boxed {\ \frac 13 (AL^2+BM^2+CN^2)\le HM^2+HN^2+HL^2<AL^2+BM^2+CN^2\ }\ .$

The condition from problem can be write: $ a^{2}+b^{2}+c^{2} < 8R^{2}$ Hence, caculate the power of $ H$ with $ (O)$, we get the rusult