In fact this problem is rather difficult. I tried for a long time to solve it until I found it as problem 13 in chapter 3 of Titu Andreescu, Dorin Andrica, 360 Problems for Mathematical Contests, Zalau 2003, published by Gil. Here is the proposed solution: We have to prove the inequality $ \frac {PA}{a^2} + \frac {PB}{b^2} + \frac {PC}{c^2}\geq \frac {1}{R}$ for every point $ P$ inside or on the boundary of triangle $ ABC$. Here, $ a = BC$, $ b = CA$, $ c = AB$ denote the sidelengths, and $ R$ denotes the circumradius of triangle $ ABC$. Proof. Let X, Y, Z be the feet of the perpendiculars from the point P to the sides BC, CA, AB of triangle ABC. The key to the solution is the inequality $ a\cdot PA\geq c\cdot PY + b\cdot PZ$. This can be proved in several ways; one very nice proof was given by Nikolaos Dergiades in his paper Signed Distances and the Erdös-Mordell Inequality, Forum Geometricorum 4 (2004) p. 67-68. Another one by Titu Andreescu goes as follows: Since < PYA = 90° and < PZA = 90°, the points Y and Z lie on the circle with diameter PA. In other words, the segment PA is a diameter of the circumcircle of triangle AYZ. Hence, by the Extended Law of Sines, $ YZ = PA\cdot \sin \measuredangle YAZ = PA\cdot \sin A$. If N and M are the orthogonal projections of the points Y and Z on the line PX, then the angles < YPN and < ACB are equal, since $ PY\perp CA$ and $ PN\perp CB$; hence, in the right-angled triangle PNY, we have $ YN = PY\cdot \sin \measuredangle YPN = PY\cdot \sin \measuredangle ACB = PY\cdot \sin C$. Similarly, $ ZM = PZ\cdot \sin B$. But since $ YN\perp PX$ and $ ZM\perp PX$, while $ PX\perp BC$, we have YN || BC and ZM || BC. Therefore, YN + ZM is equal to the projection of the segment YZ on the line BC. Hence, $ YZ\geq YN + ZM$, with equality if and only if YZ || BC. But now, since $ YZ = PA\cdot \sin A$, $ YN = PY\cdot \sin C$ and $ ZM = PZ\cdot \sin B$, we have $ PA\cdot \sin A \geq PY\cdot \sin C + PZ\cdot \sin B$. By the Law of Sines, $ \displaystyle \frac {a}{\sin A} = \frac {b}{\sin B} = \frac {c}{\sin C} > 0$; hence, we conclude $ a\cdot PA\geq c\cdot PY + b\cdot PZ$, with equality if and only if YZ || BC. Similarly, $ b\cdot PB\geq a\cdot PZ + c\cdot PX$ with equality if and only if ZX || CA, and $ c\cdot PC\geq b\cdot PX + a\cdot PY$ with equality if and only if XY || AB. Using these three inequalities, we obtain $ \displaystyle \frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}} = \frac {1}{a^{3}}\cdot a\cdot PA + \frac {1}{b^{3}}\cdot b\cdot PB + \frac {1}{c^{3}}\cdot c\cdot PC$ $ \displaystyle \geq \frac {1}{a^{3}}\cdot \left( c\cdot PY + b\cdot PZ\right) + \frac {1}{b^{3}}\cdot \left( a\cdot PZ + c\cdot PX\right) + \frac {1}{c^{3}}\cdot \left( b\cdot PX + a\cdot PY\right)$ $ \displaystyle = \left( \frac {b}{c^{3}} + \frac {c}{b^{3}}\right) \cdot PX + \left( \frac {c}{a^{3}} + \frac {a}{c^{3}}\right) \cdot PY + \left( \frac {a}{b^{3}} + \frac {b}{a^{3}}\right) \cdot PZ$. Now, the AM-GM inequality gives (1) $ \displaystyle \frac {u}{v^{3}} + \frac {v}{u^{3}}\geq \frac {2}{uv}$ for positive u and v; hence, $ \displaystyle \frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}} \geq \left( \frac {b}{c^{3}} + \frac {c}{b^{3}}\right) \cdot PX + \left( \frac {c}{a^{3}} + \frac {a}{c^{3}}\right) \cdot PY + \left( \frac {a}{b^{3}} + \frac {b}{a^{3}}\right) \cdot PZ$ $ \displaystyle \geq \frac {2}{bc}\cdot PX + \frac {2}{ca}\cdot PY + \frac {2}{ab}\cdot PZ$ (by (1)) $ \displaystyle = \frac {2}{abc}\cdot \left( a\cdot PX + b\cdot PY + c\cdot PZ\right)$. But $ a\cdot PX + b\cdot PY + c\cdot PZ = 2S$, where S is the area of triangle ABC. Thus, $ \displaystyle \frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}}\geq \frac {2}{abc}\cdot 2S = \frac {4S}{abc} = \frac {1}{R}$. This proves your inequality. In order to see when equality holds, we note that we have used the three inequalities $ a\cdot PA\geq c\cdot PY + b\cdot PZ$, $ b\cdot PB\geq a\cdot PZ + c\cdot PX$ and $ c\cdot PC\geq b\cdot PX + a\cdot PY$, and then the AM-GM inequality for the pairs of numbers $ \displaystyle \left( \frac {b}{c^{3}};\;\frac {c}{b^{3}}\right)$, $ \displaystyle \left( \frac {c}{a^{3}};\;\frac {a}{c^{3}}\right)$, $ \displaystyle \left( \frac {a}{b^{3}};\;\frac {b}{a^{3}}\right)$. In the first three inequalities $ a\cdot PA\geq c\cdot PY + b\cdot PZ$, $ b\cdot PB\geq a\cdot PZ + c\cdot PX$ and $ c\cdot PC\geq b\cdot PX + a\cdot PY$, equality holds simultaneously only if YZ || BC, ZX || CA and XY || AB, i. e. if the point P is the circumcenter of triangle ABC. In the AM-GM inequality for the pairs of numbers $ \displaystyle \left( \frac {b}{c^{3}};\;\frac {c}{b^{3}}\right)$, $ \displaystyle \left( \frac {c}{a^{3}};\;\frac {a}{c^{3}}\right)$, $ \displaystyle \left( \frac {a}{b^{3}};\;\frac {b}{a^{3}}\right)$, equality simultaneously holds only if a = b = c, i. e. if the triangle ABC is equilateral. Hence your inequality $ \frac {PA}{a^2} + \frac {PB}{b^2} + \frac {PC}{c^2}\geq \frac {1}{R}$ turns into an equality only if the triangle ABC is equilateral and the point P is its center. Darij

Oh my! No wonder I tried hours and hours on this and never got anywhere...

In post #2, I proved that the inequality $\frac{PA}{a^{2}}+\frac{PB}{b^{2}}+\frac{PC}{c^{2}}\geq\frac{1}{R}$ holds for every point P inside or on the boundary of triangle ABC. But this inequality actually is true for all points P in the plane of triangle ABC, including points outside of triangle ABC. Here is a proof of this: EDIT: The below proof is long and ugly. For a nice proof, see http://www.mathlinks.ro/Forum/viewtopic.php?t=63462 . It will be already enough to prove the following general fact: Internal Minimum Theorem. Let ABC be a triangle. If x, y, z, w are four nonnegative reals such that the inequality $x\cdot PA+y\cdot PB+z\cdot PC\geq w$ holds for every point P inside or on the boundary of triangle ABC, then this inequality $x\cdot PA+y\cdot PB+z\cdot PC\geq w$ holds for all points P in the plane of triangle ABC. Proof. Let P be a point in the plane of triangle ABC. The line BC subdivides the plane into two halfplanes. Let $\pi_{A}$ be the one which contains the point A (it should be a closed halfplane, i. e. it should include the line BC itself). Similarly, define the halfplanes $\pi_{B}$ and $\pi_{C}$. If the point P lies inside all three halfplanes $\pi_{A}$, $\pi_{B}$ and $\pi_{C}$, then this point P lies in the intersection of these three halfplanes; but the intersection of these three halfplanes is the interior and the boundary of triangle ABC, and thus, the point P must lie inside or on the boundary of triangle ABC, and hence, according to the assumption, the inequality $x\cdot PA+y\cdot PB+z\cdot PC\geq w$ holds. Hence, it is enough to consider only the case when the point P does not lie inside all of the three halfplanes $\pi_{A}$, $\pi_{B}$ and $\pi_{C}$. We can WLOG assume that the point P lies outside the halfplane $\pi_{A}$. Since the halfplane $\pi_{A}$ is the halfplane with respect to the line BC which contains the point A, saying that the point P lies outside the halfplane $\pi_{A}$ is equivalent to saying that the points A and P lie in different halfplanes with respect to the line BC. Thus, if R is the orthogonal projection of the point P on the line BC, then < PRA > 90°. Hence, in the triangle ARP, the angle < PRA is obtuse. Now, if a triangle has an obtuse angle, then this obtuse angle is the greatest angle of the triangle. Thus, the angle < PRA is the greatest angle of triangle ARP. Since, in a triangle, the greater angle lies opposite to the longer side, it thus follows that the side PA is the longest side of triangle ARP. In particular, this means that $PA\geq RA$. Also, since in a right-angled triangle, the hypotenuse is always $\geq$ to the catet (with equality for degenerate triangles), we have $PB\geq RB$ and $PC\geq RC$ in the right-angled triangles PRB and PRC, respectively. So altogether, the inequalities $PA\geq RA$, $PB\geq RB$ and $PC\geq RC$ yield $x\cdot PA+y\cdot PB+z\cdot PC\geq x\cdot RA+y\cdot RB+z\cdot RC$. Hence, in order to prove the inequality $x\cdot PA+y\cdot PB+z\cdot PC\geq w$, it is enough to verify the inequality $x\cdot RA+y\cdot RB+z\cdot RC\geq w$. Now, we know that the point R lies on the line BC. If this point R lies inside the segment BC, then we are done (in fact, in this case, the point R lies in the interior or on the boundary of triangle ABC, and thus, the inequality $x\cdot RA+y\cdot RB+z\cdot RC\geq w$ is true by the condition of the theorem). The only case we have to worry about is when the point R lies on an extension of the segment BC. We can WLOG assume that the point R lies on the extension of the segment BC beyound B. We distinguish between two cases: Case 1. The angle < ABC is acute or right. Case 2. The angle < ABC is obtuse. First consider the Case 1. In this case, the angle < ABC is acute or right. Then, the angle < ABR = 180° - < ABC must be obtuse or right. But if a triangle has an obtuse or right angle, then this angle is the greatest angle of the triangle. Thus, the angle < ABR, being obtuse or right, must be the greatest angle of triangle ARB. Now, the greater angle of a triangle lies opposite to the longer side; thus, the side RA of triangle ARB must be the longest side of this triangle. In particular, this yields $RA\geq BA$. Also, we trivially have $RB\geq BB$ (since BB = 0) and $RC\geq BC$ (since the point R lies on the extension of the segment BC beyound B). Therefore, $x\cdot RA+y\cdot RB+z\cdot RC\geq x\cdot BA+y\cdot BB+z\cdot BC$; but since the point B lies inside or on the boundary of triangle ABC (in fact, it lies very much on the boundary ), by the condition of the theorem we have $x\cdot BA+y\cdot BB+z\cdot BC\geq w$, and thus $x\cdot RA+y\cdot RB+z\cdot RC\geq w$, what completes our proof. Thus, the Case 1 is handled. Remains to consider the Case 2. In this case, the angle < ABC must be obtuse. Thus, the orthogonal projection W of the point R on the line AB must lie on the ray BA. Since the point R lies on the extension of the segment BC beyound B, the points C and R lie in different halfplanes with respect to the line AB, and thus < CWR > 90°. In other words, in the triangle CWR, the angle < CWR is obtuse. Now, if a triangle has an obtuse angle, then this obtuse angle is the greatest angle of the triangle. Thus, the angle < CWR is the greatest angle of triangle CWR. Since, in a triangle, the greater angle lies opposite to the longer side, it thus follows that the side RC is the longest side of triangle CWR. In particular, this means that $RC\geq WC$. Also, since in a right-angled triangle, the hypotenuse is always $\geq$ to the catet (with equality for degenerate triangles), we have $RB\geq WB$ and $RA\geq WA$ in the right-angled triangles RWB and RWA, respectively. Thus, $x\cdot RA+y\cdot RB+z\cdot RC\geq x\cdot WA+y\cdot WB+z\cdot WC$. Hence, in order to prove the inequality $x\cdot RA+y\cdot RB+z\cdot RC\geq w$, it is enough to establish the inequality $x\cdot WA+y\cdot WB+z\cdot WC\geq w$. Now, again, this inequality follows from the assumption of the theorem if the point W lies inside the segment AB (in fact, in this case, the point W lies inside or on the boundary of triangle ABC). The interesting case occurs when the point W lies on the extension of the segment AB beyound A (it cannot lie on the extension of the segment AB beyound B, since it lies on the ray BA). In this case, we can note that, since the angle < ABC is obtuse, and a triangle cannot have two obtuse angles, the angle < CAB must be acute or right; thus, we can prove the inequality $x\cdot WA+y\cdot WB+z\cdot WC\geq w$ for the point W in absolutely the same way as we proved the inequality $x\cdot RA+y\cdot RB+z\cdot RC\geq w$ for the point R in the Case 1 (i. e., by showing that $WC\geq AC$, $WA\geq AA$ and $WB\geq AB$). This completes the proof of the Internal Minimum Theorem, and thus the problem is solved. Don't you find this proof nice? Well, I am sure it has to distinguish a less amount of cases than the proposed solution of an average DeMO (German NMO) problem... Anyway, simplifications are always welcome! Darij

I had found another solution for this problem EDIT: The solution is wrong. Please forgive me I had mixed up the Erdoss Mordell inequality .

Thank you, Darij and Rockymarcianocosmin99, for your effort to offer the solution of the nice problem !. I consider it like a present for the new year. Once again thanks a lot !

Very sorry to revive this and I do not want to point out sucha silly error but for the sake of complete flawlessness in Darij's post number 2, he had written $ c PC\ge a PY + b PX$ where $ PC$ must be replaced by $ PZ$. Thank you Darij for such wonderful posts.....

Agr_94_Math wrote: Very sorry to revive this and I do not want to point out sucha silly error but for the sake of complete flawlessness in Darij's post number 2, he had written $ c PC\ge a PY + b PX$ where $ PC$ must be replaced by $ PZ$. Thanks for reading my post closely, but I don't see a mistake here. It should be $ PC$ on the left hand side. Maybe you are talking about a different equation? darij

Very sorry. I misread another equation. sorry for distrurbing you Darij.

This proof is quite inspired by Erdos Mordell Inequality. Let $A', B', C'$ be the projection of $P$ with respect to $BC, AC, AB$. $\textbf{Claim 01.}$ $AP \cdot BC \ge AB \cdot PB' + AC \cdot PC'$. $\textit{Proof.}$ We'll first prove that \[ AP \cdot BC \ge AB \cdot PC' + AC \cdot PB'. \]Notice that \[ BC (AP + PA') \ge BC \cdot h_a = 2[ABC] = BC \cdot PA' + AB \cdot PC' + AC \cdot PB' \]where $h_a$ denotes the $A$-altitude of $\triangle ABC$. Now, consider $P'$ such that $P'$ is the reflection of $P$ with respect to the angle bisector of $\angle BAC$. Let the projection of $P'$ on $AB$ and $AC$ be $C_1$ and $B_1$ respectively. Therefore, $AC_1 = AB'$ and $AB_1 = AC'$. Applying the above theorem on $P'$ in $\triangle ABC$. We then, have \[ AP' \cdot BC \ge AC \cdot P'B_1 + AB \cdot P'C_1 \]This is equivalent to $AP \cdot BC \ge AC \cdot PC' + AB \cdot PB' $ which is what we wanted. Therefore, we have that \[ \frac{AP}{BC^2} \ge \frac{AB}{BC^3} \cdot PB' + \frac{AC}{BC^3} \cdot PC' \]Sum cyclically, and to simplify notations, let $AB = c, BC = a, AC = b$. Therefore \begin{align*} LHS = \sum_{cyc} \frac{AP}{BC^2} &\ge \sum_{cyc} PC' \left( \frac{b}{a^3} + \frac{a}{b^3} \right) \\ &\ge \sum_{cyc} PC' \frac{2}{ab} \\ &= \frac{2}{abc} \left( \sum_{cyc} c \cdot PC' \right) \\ &= \frac{2}{abc} \cdot 2A \\ &= \frac{4A}{abc} \\ &= \frac{1}{R} \end{align*}where $A$ denotes the area of $\triangle ABC$.

Let $X, Y, Z$ be the feet of altitudes from $P$ onto $\overline{BC}, \overline{CA}, \overline{AB}$. By the Erdos-Mordell Inequality (or the key lemma thereof), $$PA \sin A \geq PY \sin C + PZ \sin B.$$Using the law of sines and suitable inequalities, $$\sum_{\mathrm{cyc}} \frac{PA}{a^2} \geq \sum_{\mathrm{cyc}} PX\left(\frac b{c^3} + \frac c{b^3}\right) \geq \sum_{\mathrm{cyc}} \frac{2PX}{bc}$$as we have the inequality $$\frac b{c^3} + \frac c{b^3} \geq \frac 2{bc} \iff \frac{(b^2-c^2)^2}{bc} \geq 0.$$But $$\sum_{\mathrm{cyc}} \frac{2PX}{bc} = \frac 1{abc}(2a \cdot PX + 2b\cdot PY + 2c \cdot PZ) = \frac{4[ABC]}{abc} = \frac 1R$$by definition, so we are done.

Very interesting!Thanks.....

Let $D,E$ and $F$ be the projection of $P$ on sides $BC, CA$ and $AB$, respectively. I will be using the following 3 inequalities \[PA\ge \frac{c}{a}\cdot PE+\frac{b}{a}\cdot PF\]\[PB\ge \frac{a}{b}\cdot PF+\frac{c}{b}\cdot PD\]\[PC\ge \frac{b}{c}\cdot PD+\frac{a}{c}\cdot PE\]where $a,b,c$ are the sides of triangle $ABC$. Using the law of sines, we can reduce the inequality to \[\frac{PA}{\sin ^2A}+\frac{PB}{\sin ^2B}+\frac{PC}{\sin ^2C}\ge 4R\]Finally, using the inequalities enumerated earlier we have \[\sum_{cyc}\frac{PA}{\sin ^2A}=\sum_{cyc}PD\cdot \left(\frac{c}{b\cdot \sin ^2B}+\frac{b}{c\cdot \sin ^2C}\right)\overset{AM-GM}{\ge} \sum_{cyc}\frac{2PD}{\sin B\sin C}=8R^2\sum_{cyc} \frac{PD}{bc}\]But \[\frac{1}{2} \sum_{cyc} \frac{PD}{bc}=\sum_{cyc} \frac{\frac{PD\cdot BC}{2}}{abc}=\frac{[ABC]}{abc}=\frac{1}{4R}\]and the conclusion follows.

By our key lemma in the proof of Erdos-Mordell, \[PA\sin\angle A\ge PE\sin\angle C+PF\sin\angle B.\]Dividing by $a^2\sin\angle A$ gives, by law of sines, \[\frac{PA}{a^2}\ge PE\cdot\frac c{a^3}+PF\cdot\frac b{a^3}.\]Summing over all vertices gives \begin{align*} \frac{AP}{a^2}+\frac{BP}{b^2}+\frac{CP}{c^2} &\ge PD\left(\frac b{c^3}+\frac c{b^3}\right)+PE\left(\frac c{a^3}+\frac a{c^3}\right)+PF\left(\frac a{b^3}+\frac b{a^3}\right)\\ &\ge PD\cdot\frac2{bc}+PE\cdot\frac2{ca}+PF\cdot\frac2{ab}\\ &=\frac 2{abc}\left(PD\cdot a+PE\cdot b+PF\cdot c\right)\\ &=\frac{4[ABC]}{abc}\\ &=\frac1R. \end{align*}Equality occurs for $P$ at the center of equilateral $\triangle ABC$.