It's known that for convex $f_i$ $\forall i$ if
$$ g(x) = \inf\{f_1(x_1)+f_2(x_2)|x_1+x_2=x\} $$
Then $g^{*}(y) = f^{*}(y)+f^{*}(y)$
I want to check this identity for $f_i(x)=\dfrac{1}{2}x^TP_ix$, $P_i > 0$.
First, I find $f^{*}$ for $f_i$:
$$ f_i^{*}(y) = \dfrac{1}{2}y^TP_i^{-1}y $$
Thus:
$$ g^{*}(y) = \dfrac{1}{2}y^T(P_1^{-1}+P_2^{-1})y $$
Then, I tried to found $g^{*}$ directly, but stuck:
$$ g^{*}(y) = \sup\limits_{x}\left(y^Tx-(\inf\limits_{x_1+x_2=x}\dfrac{1}{2}x_1^TP_1x_1 + \dfrac{1}{2}x_2^TP_1x_2)\right) $$
Here, first I found optimal $x_1, x_2$:
FOC from Lagrangian for "inf" problem:
$$ x_i^TP_i = \nu^T => x_1^TP_1 = x_2^TP_2 \\ x_1 + x_2 = x $$
Thus:
$$ x_1 = (I+P_2^{-1}P_1)^{-1}x\\ x_2 = (I - (I+P_2^{-1}P_1)^{-1})x $$
I don't know how to simplify quadratic form, when inserting this in equation for $g^{*}(y)$