Let $f(P)=\frac{PA}{h_b+h_c}+\frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}$ for any point $P$ in the plane of $\triangle{ABC}$. We first present a lemma, and then consider a few cases. Lemma: $f(A),f(B),f(C)\ge1$. Proof: We just prove the claim for $f(A)$. Note that if $a\ge b,c$, then \begin{align*} f(A)=\frac{AB}{h_a+h_c}+\frac{AC}{h_a+h_b} &= \frac{c}{2[ABC]\left(\frac1a+\frac1c\right)}+\frac{b}{2[ABC]\left(\frac1a+\frac1b\right)}\\ &= \frac{a}{2[ABC]}\left(\frac{c^2}{a+c}+\frac{b^2}{a+b}\right)\ge \frac{a}{2[ABC]}\left(\frac{c^2}{a+a}+\frac{b^2}{a+a}\right)\\ &= \frac{b^2+c^2}{4[ABC]} \ge \frac{2bc}{4[ABC]} \ge \frac{4[ABC]}{4[ABC]} = 1. \end{align*}If $a$ is not the maximum, and, WLOG, $c\ge a,b$, then \begin{align*} f(A)=\frac{AB}{h_a+h_c}+\frac{AC}{h_a+h_b} &= \frac{c}{2[ABC]\left(\frac1a+\frac1c\right)}+\frac{b}{2[ABC]\left(\frac1a+\frac1b\right)}\\ &= \frac{a}{2[ABC]}\left(\frac{c^2}{a+c}+\frac{b^2}{a+b}\right)\ge \frac{a}{2[ABC]}\left(\frac{(b+c)^2}{2a+b+c}\right)\\ &= \frac{a}{2[ABC]}\left(b+\frac{2bc+c^2-2ab-bc}{2a+b+c}\right) = \frac{a}{2[ABC]}\left(b+\frac{2b(c-a)+c(c-b)}{2a+b+c}\right)\\ &\ge \frac{ab}{2[ABC]} \ge \frac{2[ABC]}{2[ABC]} = 1.\blacksquare \end{align*} Case 1: $\frac{1}{h_b+h_c},\frac{1}{h_a+h_c},\frac{1}{h_a+h_b}$ do not form the sides of a triangle. WLOG, we assume \[\frac{1}{h_b+h_c} \ge \frac{1}{h_a+h_c}+\frac{1}{h_a+h_b}.\]We now prove that $f(A)<f(P)$ for all $P$ inside $\triangle{ABC}$. Indeed, \begin{align*} f(P) &= \frac{PA}{h_b+h_c}+\frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b} \ge \frac{PA}{h_a+h_c}+\frac{PA}{h_a+h_b}+\frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}\\ &= \frac{PA+PB}{h_a+h_c}+\frac{PA+PC}{h_a+h_b} > \frac{AA}{h_b+h_c}+\frac{AB}{h_a+h_c}+\frac{AC}{h_a+h_b} = f(A). \end{align*}But by the lemma, $f(A)\ge1$, so we're done with this case. Case 2: $\frac{1}{h_b+h_c},\frac{1}{h_a+h_c},\frac{1}{h_a+h_b}$ do form the sides of a triangle. Construct point $A'$ not on the same side of $BC$ as point $A$ such that \[A'B:A'C:BC = \frac{1}{h_a+h_b}:\frac{1}{h_a+h_c}:\frac{1}{h_b+h_c}.\]Let $\alpha=\angle{CA'B}$, $\beta=\angle{A'BC}$, and $\gamma=\angle{BCA'}$. Note that at most one of $\alpha+A$, $\beta+B$, $\gamma+C$ is not less than $\pi$, or else, WLOG, $2\pi=\alpha+\beta+\gamma+A+B+C>(\alpha+A)+(\beta+B)\ge2\pi$, a contradiction. There are two subcases to consider. Subcase 2.1: WLOG, $\gamma+C\ge\pi>\alpha+A,\beta+B$. By Ptolemy's Inequality on the four points $A',B,P,C$, \[PB\cdot A'C+PC\cdot A'B \ge PA'\cdot BC \Longleftrightarrow \frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}\ge\frac{PA'}{h_b+h_c}.\]Thus \[f(P)=\frac{PA}{h_b+h_c}+\frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}\ge\frac{PA+PA'}{h_b+h_c}.\]Since $\gamma+C=\angle{ACB}+\angle{BCA'}\ge\pi$ and $P$ is inside $\triangle{ABC}$, we can extend $AC$ to hit segment $PA'$ at point $X$. Then the Triangle Inequality gives us $AP+PX\ge AX=AC+CX$ and $CX+A'X\ge A'C$, so adding, we have $AP+PX+A'X=AP+A'P\ge AC+A'C$. This implies that \[f(P)\ge\frac{PA+PA'}{h_b+h_c}\ge\frac{CA+CA'}{h_b+h_c},\]with equality if and only if $P=C$. This subcase now follows from the lemma. Subcase 2.2: $\pi>\alpha+A,\beta+B,\gamma+C$. By Ptolemy's Inequality on quadrilateral $A'BPC$, \[PB\cdot A'C+PC\cdot A'B \ge PA'\cdot BC \Longleftrightarrow \frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}\ge\frac{PA'}{h_b+h_c}.\]Thus by the Triangle Inequality, \[f(P)=\frac{PA}{h_b+h_c}+\frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}\ge\frac{PA+PA'}{h_b+h_c}\ge\frac{AA'}{h_b+h_c},\]with equality if and only if $P$ is on the circumcircle of $\triangle{A'BC}$ and on line $AA'$. Simple angle chasing yields that $P$ is on lines $BB',CC'$ and the circumcircles of $\triangle{B'CA},\triangle{C'AB}$ as well, where $B',C'$ are defined analogously to $A$, i.e. \[B'C:B'A:CA = \frac{1}{h_b+h_c}:\frac{1}{h_a+h_b}:\frac{1}{h_a+h_c}\]and \[C'A:C'B:AB = \frac{1}{h_a+h_c}:\frac{1}{h_b+h_c}:\frac{1}{h_a+h_b}.\]Now we know where the optimal $P$ is, and that for this $P$, \[f(P)=\frac{PA}{h_b+h_c}+\frac{PB}{h_a+h_c}+\frac{PC}{h_a+h_b}=\frac{AA'}{h_b+h_c}=\frac{BB'}{h_a+h_c}=\frac{CC'}{h_a+h_b}.\]Now how to complete the problem...

First of all, let us note that $2S=ah_a=bh_b=ch_c=a\cdot PD+b\cdot PE+c\cdot PF$ where $S$ is the area of the triangle $ABC$. Also, from the lemmas above, we have $PA\ge \frac{(b+c)(PE+PF)}{2a}$ and similar. Now, we have $\sum\frac{PA}{h_b+h_c}=\sum\frac{PA}{\frac{bh_b}{b}+\frac{ch_c}{c}}=\sum\frac{PA}{\frac{2S}{b}+\frac{2S}{c}}=\sum\frac{bc\cdot PA}{2S(b+c)}$. Using the second inequality gives $\sum \frac{bc\cdot PA}{2S(b+c)}\ge\sum\frac{bc\cdot \frac{(b+c)(PE+PF)}{2a}}{2S(b+c)}$ $=\sum \frac{bc(PE+PF)}{4aS}=\frac{\sum a^2b^2(PD+PE)}{4abcS}$. The first equality gives $\frac{\sum a^2b^2(PD+PE)}{4abcS}=\frac{a^2b^2(PD+PE)+b^2c^2(PE+PF)+c^2a^2(PF+PD)}{2abc(a\cdot PD+b\cdot PE+c\cdot PF)}$. However, upon expanding, the last expression is equivalent to $\frac{(a^2b^2+c^2a^2)PD+(b^2c^2+a^2b^2)PE+(b^2c^2+c^2a^2)PF}{2a^2bc\cdot PD+2ab^2c\cdot PE+2abc^2\cdot PF}$, which is indeed greater than 1 by AM-GM Inequality. We are done. $\blacksquare$

Notice that $h_b=\frac{2[ABC]}{b}$, etc. Hence the inequality is $$\sum_{cyc}\frac{PA}{\frac{1}{b}+\frac{1}{c}}\ge 2[ABC]$$$$\sum_{cyc}PA\frac{bc}{b+c}\ge 2[ABC]$$Let $x=d(P, AC), y=d(P, AB), z=d(P, BC)$. This implies that $$2[ABC]=xb+yc+za$$By the key lemma in the proof of Erdos-Mordell (I think the name is Mordell lemma) $$PAa\ge xb+yc, xc+yb$$Adding yields $$PA\ge \frac{(x+y)(b+c)}{2a}$$$$\sum_{cyc}PA\frac{bc}{b+c}\ge \sum_{cyc}\frac{(x+y)bc}{2a}$$But notice that $$\sum_{cyc}\frac{xbc+ybc}{2a}=\frac{1}{2}\sum_{cyc}xb\frac{c}{a}+yc\frac{b}{a}$$But $$\sum_{cyc}xb\left(\frac{c}{a}+\frac{a}{c}\right)\ge 2\sum_{cyc}xb$$by AM-GM. This directly implies the result.

Drop perpendiculars from $P$ onto $BC, AC, AB$ called $D, E, F$ respectively. Let $x = PD$, $y=PE$, and $Z=PF$. Construct $S$ such that $BCSP$ is a parallelogram in that order and $T$ such that $BATP$ is a parallelogram in that order. By Pappus’s Theorem, $[BCSP] + [BATP] = [ACST]$. Note that $[ACST] \le AC \times AT = b \cdot BP$. This becomes $b \cdot BP \ge cz + ax$. Now, let $A’$ and $C’$ be the reflections of $A$ and $C$ about the angle bisector of $\angle BAC$. Now, construct $S’$ such that $BA’S’P$ is a parallelogram in that order and construct $T’$ such that $BC’T’P$ is a parallelogram in that order. By Pappus’s Theorem, we have that $[BA’S’P] + [BC’T’P] = [C’A’S’T’]$. Note that $[BA’S’P] = x \cdot BA’ = cx$ and $[BC’T’P] = BC’ \cdot z = az$. Also, $[C’A’S’T’] \le C’A’ \cdot C’T’ = b \cdot BP$. So, $b \cdot BP \ge cx + az$. Generating all cyclic permutations of the above inequalities, we get that \[a \cdot AP \ge by + cz\]\[b \cdot BP \ge cz + ax\]\[c \cdot CP \ge ax + by\]and \[a \cdot AP \ge bz + cy\]\[b \cdot BP \ge cx + az\]\[c \cdot CP \ge ay + bx.\]These are fairly well-known and often called Mordell’s Lemma, but I reproved them here. The result we want to prove is that \[\sum_{cyc} \frac{PA}{\frac{2K}{b} + \frac{2K}{c}} \ge 1,\]which is equivalent to \[2\sum_{cyc}\frac{PA \cdot abc}{a(b + c)} \ge 4K.\]Using the two different inequalities on $PA \cdot a$ that we have, we get that \begin{align*} 2\sum_{cyc}\frac{PA \cdot abc}{a(b + c)} &\ge \sum_{cyc} \frac{(by + cz)bc}{a(b + c)} + \sum_{cyc} \frac{(bz + cy)bc}{a(b + c)} \\ &= \sum_{cyc} \frac{(y + z)bc}{a} \\ &= \sum_{cyc} ax\left(\frac{b}{c} + \frac{c}{b}\right)\\ &\ge 4K. \end{align*}So, we’re done.