The ground state of Helium atom is a state in which the space part of the wavefunction is symmetric and the spin part is antisymmetric under the exchange of the electrons. Therefore, the ground state is a spin singlet state. Theoretically, can we show that the singlet state has a lower energy than the triplet state? I think to derive it from the perturbation theory calculation, we need to show that the exchange integral $$J= \int d^3 x_1 \int d^3x_2 \psi_{100}(\vec{x}_1)\psi_{nlm}(\vec{x}_2)\frac{e^2}{r_{12}}\psi^*_{100}(\vec{x}_2)\psi^*_{nlm}(\vec{x}_1)$$ is positive. This is because the energy associated with the spin singlet and triplet states are respectively $I-J$ and $I+J$ where $I$ is the direct integral which is always positive.
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First of all, note that the simple description of helium atom, where spin-orbital coupling is neglected, allows to solve for spinor and spatial part of the wavefunction separately. Moreover, for each symmetric and asymmetric spatial eigenfunction there exist corresponding opposite-type-symmetry spinor values, which means that none of the spatial eigenfunctions are forbidden by Pauli principle.
Now, what remains to be shown is that lowest energy spatial wavefunction is symmetric. This is true by Courant nodal domain theorem$^\dagger$: lowest energy eigenfunction must have no nodes. Since an antisymmetric in electron positions function must have a node at the locus of $e^--e^-$ collision, lowest energy eigenfunction must indeed be symmetric.
$^\dagger$ Courant, Hilbert, "Methods of Mathematical Physics", vol. 1, §VI.6 "Nodes of eigenfunctions"
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To make the spin triplet state while satisfying the exclusion principle, you would have to change one of the spatial wavefunctions to $n=2$. That requires an amount of energy equal to $E_2-E_1$.