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Question:

Consider a triangle $\Delta ABC$ with altitudes $h_{a}$, $h_{b}$ and $h_{c}$, where $AB=c$, $BC=a$ and $AC=b$. Show that for any $P$

$$\sqrt{PA+PB}+\sqrt{PB+PC}+\sqrt{PA+PC}\ge 2\sqrt{h_{a}+h_{b}+h_{c}}$$

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My try: the inequality is equivalent to

$$(PA+PB+PC)+\sum_{cyc}\sqrt{(PA+PB)(PB+PC)}\ge 2(h_{a}+h_{b}+h_{c})$$

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math110
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  • To confirm: $h_a$, $h_b$, and $h_c$ are the altitudes in $\triangle ABC$, and are not altitudes from $P$ in the subtriangles? – Peter Woolfitt Aug 01 '14 at 06:08
  • @PeterWoolfitt,Yes,Thank you – math110 Aug 02 '14 at 03:01
  • @chinamath Did you try an analytic approach? Say, A is the origin and $B=(B_1, B_2)$, $C = (C_1, C_2)$. Since a triangle is convex you can describe it by two parameters $s$, $t > 0$ and every point $Q$ inside the triangle is given as $Q = s \cdot \vec{AB} + t \cdot \vec{AC}$ with $s + t = 1$. The sum of the heights can be expressed through $a$, $b$, $c$. The only thing left to do is find all the local and global minima of the expression on the left-hand sinde within the triangle. It is cumbersome and not elegant at all, but it should get the job done. – RogueDodecahedron Sep 09 '14 at 12:40
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    @chinamath have you an idea when the sign of equality holds ? – arunoruto Nov 20 '14 at 14:20
  • equality holds at the center of an equilateral triangle – Marconius Jul 10 '15 at 20:51
  • Things one can assume without loss of generality:

    \begin{align} A&=(0,0)\ B&=(1,0)\ C&=(x,y)\text{ with $y>0$}\ P&=(p,q) \text{ lies in $\triangle ABC$} \end{align}

    The last one in particular implies $0\leq q \leq y$ and $py-qx\geq 0$.

    It's possible to reduce the statement inequality to a polynomial inequality in $PA,PB,PC, h_a,h_b$ and $h_c$, but it's a rather involved expression and I'm not sure it's of much use.

     – Fimpellizzeri Jun 05 '17 at 01:51

1 Answers1

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I noticed that the LHS is dependent on $P$ but the RHS isn't. So why not try the following approach: Find out the $P$ for which the LHS is minimum , and prove the inequality. Then we are done!

For any given triangle $ABC$, the minimum value of LHS will occur if the $P$ is the Circumcentre. i.e. $PA=PB=PC=R\ (circumradius)$. The inequality reduces to
$$ 9R\ge 2(h_a+h_b+h_c)$$ Now we know $R,h_a,h_b,h_c$ in terms of the sides and angles. Can you proceed from here?

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