Added: I have created some code that verifies the inequality is true for Dynkin diagrams of type $A_n,B_n,C_n,D_n$ for $n\leq 10$, and it has verified the special type $E_6,E_7,E_8,F_4,G_2$. I'll post an answer here once I have a formal proof for general $A,B,C,D$ types.
Let $X=G/P$ be a flag variety, i.e. $G$ is a connected reductive linear algebraic group and $P\subseteq G$ is a parabolic subgroup (everything is over an algebraically closed field of characteristic $0$).
Then the anticanonical divisor of $X$ can be written as $$-K_X = \sum_{\alpha\in S\setminus I} b_\alpha D_\alpha$$ where $S$ is the simple root set of $G$, and $I$ is the simple root set for $P$; here the $D_\alpha$ are the Schubert divisors. These coefficients $b_\alpha$ are positive integers. My question is about how these coefficients relate to the dimension of the flag variety, namely:
Do we have the inequality $$\sum_{\alpha\in S\setminus I} (b_\alpha-1) \leq \dim(X) ~?$$
I have tried a few examples, e.g. $X=SL_n/B$ and projective spaces written as $SL_n/P$ where $P$ is a maximal parabolic, and the answer is yes (with equality for the projective spaces).
Added: we have $b_\alpha=\langle \sum_{\beta\in R^+\setminus R_I} \beta,\alpha^\vee\rangle$ where $R^+$ is the positive root set for $G$, and $R_I$ is the root set generated by $I$; this pairing is the root/dual root pairing. Thus, $$\sum_{\alpha\in S\setminus I} b_\alpha = \sum_{\alpha\in S\setminus I}\sum_{\beta\in R^+\setminus R_I} \langle \beta,\alpha^\vee\rangle.$$
Added: since $\dim(X)=\#(R^+\setminus R_I)$, we can reformulate my question as:
Reformulation of my question: do we have the inequality $$\sum_{\alpha\in S\setminus I}\sum_{\beta\in R^+\setminus R_I} \langle \beta,\alpha^\vee\rangle \leq \#(R^+\setminus R_I)+\#(S\setminus I) ~?$$
I don't believe this is true if I have done my calculations correctly although it will hold for $G$ of type $A_n,D_n,E_n$. It works in those cases as $\langle\beta,\alpha^\vee\rangle$ is one of $-1,0,1,2$ and is only $2$ when $\beta = \alpha$. This comes down to facts about the length of root strings in these types. Thus the sum cannot be more than the number of roots in our nilradical ($\#(R^+/R_I) $ in your notation) + the number of simple roots in the nilradical ($\#(S/I) $) and we have equality exactly when all the $\langle\beta,\alpha^\vee\rangle$ are $1$ except for each $\langle\alpha,\alpha^\vee\rangle$ (clearly not possible as soon as there is more than one simple root).
However if we look at $G_2$ we quickly find a counterexample. Choosing the maximal parabolic given by the short simple root $\alpha_1$ I obtain $b_{\alpha_1} = 15$ but the corresponding flag manifold is of dimension $5$
Edit: I think I may have made a mistake in fact. Here are my calculations for the $G_2$ case. There are two simple roots $\alpha_1$, $\alpha_2$ and the positive roots are $\alpha_1,\alpha_2, \alpha_1+\alpha_2, 2\alpha_1+\alpha_2 , 3\alpha_1+\alpha_2, 3\alpha_1+2\alpha_2$ of which only $\alpha_2$ is in $R_I$ in our example.
Then $\langle\alpha_1, \alpha_1^\vee\rangle =2$ and $\langle\alpha_2, \alpha_1^\vee\rangle =-3$ (I had originally put this last one as $-1$ but that is $\langle\alpha_1, \alpha_2^\vee\rangle$)
So we obtain $$\langle\alpha_1, \alpha_1^\vee\rangle =2$$
$$\langle\alpha_1+\alpha_2, \alpha_1^\vee\rangle =-1$$
$$\langle 2\alpha_1+\alpha_2, \alpha_1^\vee\rangle =1$$
$$\langle 3\alpha_1+\alpha_2, \alpha_1^\vee\rangle =3$$
$$\langle 3\alpha_1+2\alpha_2, \alpha_1^\vee\rangle =0$$ Which adds up to 5.