Only if the triangle $ABC$ is acute or right ! Omid Hatami wrote: In acute or right triangle $ABC$, if $L,M,N$ are midpoints of the its sides and $H$ is its orthocenter, then show that $\boxed{\ LH^{2}+MH^{2}+NH^{2}\leq\frac14\cdot \left(a^{2}+b^{2}+c^{2}\right)\ }$ Proof. It is well-known that $HA=2R|\cos A|$, $a=2R\sin A$ a.s.o. and $\Delta\equiv\sum \cos 2A=-1-4\prod\cos A\ .$ Therefore, $4\sum HL^{2}-\sum a^{2}=$ $\sum\left(4\cdot HL^{2}-a^{2}\right)=$ $\sum\left[2\left(HB^{2}+HC^{2}\right)-2a^{2}\right]=$ $4\sum HA^{2}-2\sum a^{2}=$ $16R^{2}\sum \cos^{2}A-8R^{2}\sum \sin^{2}A=$ $8R^{2}\sum (1+\cos 2A)-4R^{2}\sum (1-\cos 2A)=$ $8R^{2}(3+\Delta )-4R^{2}(3-\Delta )=$ $4R^{2}(6+2\Delta-3+\Delta)=$ $12R^{2}(1+\Delta )=$ $-48R^{2}\cos A\cos B\cos C\le 0$ $\Longrightarrow$ $\sum HL^{2}\le\frac{1}{4}\cdot\sum a^{2}\ .$ Remark. I posted long ago the following inequality : $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ ,$ where $I$ is the incenter of the triangle $ABC$. Try solve it !

A correction : $\sum IM^{2}\ge r(R+r)$

I have wondered this for a long time: what does you a.s.o stand for? Is it something along the lines of "and symmetric others"?

$\left\{\begin{array}{c}4\cdot IM^{2}=2\left(IB^{2}+IC^{2}\right)-a^{2}\\\\ IB^{2}=\frac{ac(p-b)}{p}\\\\ IC^{2}=\frac{ab(p-c)}{p}\end{array}\right\|$ $\Longrightarrow$ $4p\cdot IM^{2}=2ac(p-b)+2ab(p-c)-pa^{2}=a[c(a+c-b)+b(a+b-c)-ap]=$ $a\left[(b-c)^{2}+a(b+c)-ap\right]=a(b-c)^{2}+a^{2}(p-a)$ $\Longrightarrow$ $\left\{\begin{array}{c}4p\cdot\sum IM^{2}=\sum \left[a(b-c)^{2}\right]+\sum a^{2}(p-a)\\\\ \sum a^{2}(p-a)=4S(R+r)\end{array}\right\|$ $\Longrightarrow$ $4p\cdot \sum IM^{2}=\sum \left[a(b-c)^{2}\right]+4pr(R+r)$ $\Longrightarrow$ $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ .$

Forall $P$ on plane then $PA'^{2}+PB'^{2}+PC'^{2}\ge\frac{a^{2}+b^{2}+c^{2}}{12}\ge\frac{r(4R+r)}{3}\ge r(R+r)$ with $A',B',C'$ are midpoint of $BC,CA,AB$ resp Thus the result of Virgil Nicula is true for all $P$ on plane! And more here we can show min of $\sum PA'^{2}$ with a fix triangle $ABC$ and really it is only simple result!

Omid Hatami wrote: In triangle $ABC$, if $L,M,N$ are midpoints of $AB,AC,BC$. And $H$ is orthogonal center of triangle $ABC$, then prove that \[LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})\] i tried in complex co-ordinat and the result became $2abc+ \sum (a^2b+b^2a) \ge 0$ is this true? $a,b,c$ are co-ordinat of $A,B,C$ and $h=a+b+c$ $l=(a+b)/2$ $m=(a+c)/2$ $n=(b+c)/2$

tsemng wrote: Omid Hatami wrote: In triangle $ABC$, if $L,M,N$ are midpoints of $AB,AC,BC$. And $H$ is orthogonal center of triangle $ABC$, then prove that \[LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})\] i tried in complex co-ordinat and the result became $2abc+ \sum (a^2b+b^2a) \ge 0$ is this true? $a,b,c$ are co-ordinat of $A,B,C$ and $h=a+b+c$ $l=(a+b)/2$ $m=(a+c)/2$ $n=(b+c)/2$ Hmm, I am getting a different result. Denote by $H$, the origin; and give $A,B,C$ the affixes $u,v,w.$ So we see that $l=\frac{v+w}{2}, m=\frac{w+u}{2},n=\frac{u+v}{2}.$ Therefore $AB^2+BC^2+CA^2=(u-v)^2+(v-w)^2+(w-u)^2$ and $LP^2+MP^2+NP^2=\frac 14\left[(w+u)^2+(u+v)^2+(v+w)^2\right].$ Also, $u+v+w=0$; so that \[\begin{aligned}\frac 14\left[AB^2+BC^2+CA^2\right]&=\frac 14\left[(u-v)^2+(v-w)^2+(w-u)^2\right]\\&=\frac 14\left[2(u+v+w)^2+(u-v)^2+(v-w)^2+(w-u)^2\right]\\&=\frac 14\left[2(u^2+v^2+w^2)+(u+v)^2+(v+w)^2+(w+u)^2\right]\\&\geq \frac 14\left[(u+v)^2+(v+w)^2+(w+u)^2\right];\end{aligned}\] And we are done. $\Box$

And so what is the equality case ? Is it when the triangle is reduced to a point ? Thanks in advance.

Only if $ABC$ is acute or right. Equality holds if it is right. We use the fact that reflection of the orthogonal center with the respect to the midpoints of the triangle lay on the circumcircle. Denote $P,Q$ intersection of $HL$ with circumcircle. ($P$ is reflection of $H$ wrt $L$) Looking at power of the point $L$ wrt to the circumcircle and noticing that $LP=HL \leq LQ$ we get $\frac{AB^2}{4}=AL*BL=LQ*LP \geq HL^2$ Applying this to two other sides we get the desired result.

From Leybnits's identity it's equivalent to $ cosA \cdot cosB \cdot cosC \ge 0 $.It's true if and only if $ ABC $ is non-obtuse!

Virgil Nicula wrote: Only if the triangle $ABC$ is acute or right ! Omid Hatami wrote: In acute or right triangle $ABC$, if $L,M,N$ are midpoints of the its sides and $H$ is its orthocenter, then show that $\boxed{\ LH^{2}+MH^{2}+NH^{2}\leq\frac14\cdot \left(a^{2}+b^{2}+c^{2}\right)\ }$ Proof. It is well-known that $HA=2R|\cos A|$, $a=2R\sin A$ a.s.o. and $\Delta\equiv\sum \cos 2A=-1-4\prod\cos A\ .$ Therefore, $4\sum HL^{2}-\sum a^{2}=$ $\sum\left(4\cdot HL^{2}-a^{2}\right)=$ $\sum\left[2\left(HB^{2}+HC^{2}\right)-2a^{2}\right]=$ $4\sum HA^{2}-2\sum a^{2}=$ $16R^{2}\sum \cos^{2}A-8R^{2}\sum \sin^{2}A=$ $8R^{2}\sum (1+\cos 2A)-4R^{2}\sum (1-\cos 2A)=$ $8R^{2}(3+\Delta )-4R^{2}(3-\Delta )=$ $4R^{2}(6+2\Delta-3+\Delta)=$ $12R^{2}(1+\Delta )=$ $-48R^{2}\cos A\cos B\cos C\le 0$ $\Longrightarrow$ $\sum HL^{2}\le\frac{1}{4}\cdot\sum a^{2}\ .$ Remark. I posted long ago the following inequality : $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ ,$ where $I$ is the incenter of the triangle $ABC$. Try solve it ! Sorry,I don,t know why LH^2-a^2=2(BH^2+CH^2)-2a^2 Can you explain?

$ O $ the center of $ \odot (ABC) $ and WLOG let $\odot (ABC)$ be unit circle vectors By Euler line we get $ H=a+b+c $ $ LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2}) \Longleftrightarrow \sum_{cyc} (\frac{a+b+2c}{2})^2 \le \sum_{cyc} (\frac{a-b}{2})^2 $ We use that $ a^2=b^2=c^2=1 $ $ \Leftrightarrow -1 \ge ab+bc+ca $ By scalar product ,$ ab= cos(2C) $, $ \Leftrightarrow -1 \ge cos(2A)+cos(2B)+cos(2C) $ $ \Leftrightarrow -1 \ge 2cos(A+B)cos(A-B)+2cos^2(A+B)-1 $ $ \Leftrightarrow 0 \ge +4cos(A+B)cosAcosB $ $ \Leftrightarrow 0 \ge -4cos(C)cos(A)cos(B) $ This is true for acute or right triangle.