To simplify derivation introduce the notation:
$$
BC=a,\quad AC=b,\quad AB=c,\\
\measuredangle{CAB}=\alpha,\quad
\measuredangle{ABC}=\beta,\quad
\measuredangle{BCA}=\gamma;\\
\measuredangle{ABP}=\alpha_B,\quad
\measuredangle{ACP}=\alpha_C,\\
\measuredangle{BAP}=\beta_A,\quad
\measuredangle{BCP}=\beta_C,\\\
\measuredangle{CAP}=\gamma_A,\quad
\measuredangle{CBP}=\gamma_B.
$$
Then the initial inequality can be written as:
$$\small
\frac{a(\sin\alpha_B+\sin\alpha_C)
+b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\le1.\tag{*}
$$
which can be proved as follows:
$$\begin{align}\small
&\scriptsize\frac{a(\sin\alpha_B+\sin\alpha_C)
+b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\\
&\scriptsize=\frac{\sin\alpha(\sin\alpha_B+\sin\alpha_C)
+\sin\beta(\sin\beta_A+\sin\beta_C)+\sin\gamma(\sin\gamma_A+\sin\gamma_B)}
{\sin\alpha+\sin\beta+\sin\gamma}\tag1\\
&\scriptsize=\frac{(\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A)
+(\sin\alpha\sin\alpha_B+\sin\gamma\sin\gamma_B)
+(\sin\alpha\sin\alpha_C+\sin\beta\sin\beta_C)}
{\sin\alpha+\sin\beta+\sin\gamma}\tag2\\
&\scriptsize\le\frac{\sin\alpha+\sin\beta+\sin\gamma}{\sin\alpha+\sin\beta+\sin\gamma}=1\tag3.
\end{align}
$$
Explanation:
(1)
Substitution:
$$a=2R\sin\alpha, \quad b=2R\sin\beta, \quad c=2R\sin\gamma.$$
(2)
Rearrangement.
(3) Transformation of sine products to cosine differences and then from cosine sums to cosine products:
$$\begin{align}
&\scriptsize\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A\\
&\scriptsize=\frac{\cos(\beta-\beta_A)-\cos(\beta+\beta_A)}2
+\frac{\cos(\gamma-\gamma_A)-\cos(\gamma+\gamma_A)}2\\
&\scriptsize=\frac{\cos(\beta-\beta_A)+\cos(\gamma-\gamma_A)}2
-\frac{\cos(\beta+\beta_A)+\cos(\gamma+\gamma_A)}2\\
&\scriptsize=
\underbrace{\cos\frac{\beta+\gamma-\beta_A-\gamma_A}2}_{\ge0}
\underbrace{\cos\frac{\beta-\gamma-\beta_A+\gamma_A}2}_{\le1}
-
\underbrace{\cos\frac{\beta+\gamma+\beta_A+\gamma_A}2}_{=0}\cos\frac{\beta-\gamma+\beta_A-\gamma_A}2\tag4\\
&\scriptsize\le \cos\frac{\beta+\gamma-\alpha}2=\cos\left(\frac\pi2-\alpha\right)=\sin\alpha,
\end{align}
$$
where we used $\beta_A+\gamma_A=\alpha$.
There may arise an interesting question when does the inequality $(\text{*})$ degenerate to equality?
It is easily seen from $(4)$ that this happens if and only if the following equations hold:
$$
\begin {cases}
\beta-\beta_A=\gamma-\gamma_A\\
\gamma-\gamma_B=\alpha-\alpha_B\\
\alpha-\alpha_C=\beta-\beta_C
\end {cases},
$$
which together with
$$
\begin {cases}
\beta_A+\gamma_A=\alpha\\
\gamma_B+\alpha_B=\beta\\
\alpha_C+\beta_C=\gamma
\end {cases}
$$
implies:
$$
\begin {cases}
\beta_C=\gamma_B=\frac\pi2-\alpha\\
\gamma_A=\alpha_C=\frac\pi2-\beta\\
\alpha_B=\beta_A=\frac\pi2-\gamma
\end {cases}.
$$
Thus the equality holds if and only if the triangle $ABC$ is acute and $P$ is its circumcenter.
Note:
With the convention that the angle measures at a triangle vertex are signed and the direction towards the adjacent side is positive, the inequality (*) becomes valid for the whole plane. Correspondingly it degenerates to equality at the circumcenter of the triangle, regardless of its shape.
Let $D$ be a point on the line segment $AB$ such that $VD=\frac{VA+VB}{2}$.
(Such a point $D$ exists. The reason is as follows : Let $A'$ be a point on the half line $VA$ such that $VA'=\frac{VA+VB}{2}$. Let $B'$ be a point on the half line $VB$ such that $VB'=\frac{VA+VB}{2}$. Then, we have $VA'\gt VA$ since $$2VA\lt VA+VB\implies VA\lt \frac{VA+VB}{2}=VA'$$ Similarly, $VB'\lt VB$ since $$2VB\gt VA+VB\implies VB\gt\frac{VA+VB}{2}=VB'$$ Now on the plane $VAB$, let us consider a circle whose center is $V$ with radius $\frac{VA+VB}{2}$. We see that $A',B'$ are on the circle. Since $VA'\gt VA$ and $VB'\lt VB$, there is a point $D$ on the line segment $AB$ such that $D$ is on the circle. For such a point $D$, we have $VD=\frac{VA+VB}{2}$.)
Let $E$ be a point on the line segment $VD$ such that $VE=\frac 23VD$.
Let $F$ be a point on the line segment $VC$ such that $VF=\frac 13VC$.
Let $G$ be a point on the half line $VD$ such that $VG=VE+VF$.
Let $H$ be a point on the half line $VC$ such that $VH=VE+VF$.
Now, since $VD=\frac{VA+VB}{2}$, we have
$$\frac{VA+VB+VC}{3}=\frac{2VD+VC}{3}=VE+VF$$
We have $VG\gt VD$ since
$$\begin{align}&VC-VA+VC-VB\gt 0
\\&\implies VC\gt \frac{VA+VB}{2}
\\&\implies VC\gt VD
\\&\implies \frac 13VC\gt \frac 13VD
\\&\implies VF\gt ED
\\&\implies VF+VE\gt ED+VE
\\&\implies VG\gt VD\end{align}$$
Also, we have $VC\gt VH$ since
$$\begin{align}\frac 23VC\gt\frac 23VD&\implies FC\gt VE
\\&\implies FC+VF\gt VE+VF
\\&\implies VC\gt VH\end{align}$$
Now, on the plane $VCD$, let us consider a circle whose center is $V$ with radius $VE+VF$.
$G,H$ are on the circle with $VG\gt VD$ and $VH\lt VC$.
So, there is a point $P$ on the line segment $CD$ such that $P$ is on the circle.
For such a point $P$, we have $$VP=VE+VF=\frac{VA+VB+VC}{3}. \blacksquare$$
Best Answer
WLOG the hypotenuse is 1. Let the legs be $a,b$. Denote $ab = x$. $$(1-a)(1-b) = \frac{a^2b^2}{1+a+b+ab} \le \frac{a^2b^2}{1+2\sqrt {ab}+ab} = \left(\frac{x}{1+\sqrt x}\right)^2$$ Now, consider: $$\frac{x}{1+\sqrt x} = \sqrt x - \frac{\sqrt x}{\sqrt x+1} = \sqrt x -1 + \frac 1{\sqrt x+1} = \sqrt x+1 + \frac 1{\sqrt x+1} -2$$ The function $f(y) = y+\frac 1y$ is increasing for $y>1$, so the expression attains max value when $\sqrt x$ is maximum. Verify that this happens when $a=b$ as: $$\sqrt{ab} \le \sqrt{(a^2+b^2)/2} = \frac 1{\sqrt 2}$$by AM-GM. Equality when $a=b$.
Claim: $\forall y>1$, $y+1/y$ is strictly increasing.
Proof: Let $m>n>1$. Observe that $$\left(m+\frac 1m\right)-\left(n+\frac 1n\right)$$ $$ = (m-n)+\left(\frac 1m - \frac 1n\right)$$ $$= (m-n)+\left(\frac {n-m}{mn}\right)$$ $$= (m-n)\left(1 - \frac 1{mn}\right) >0$$