Intuitively, why is the energy decrease of the singlet state three times larger than the energy increase of the triplet state (hyperfine correction)? Following Griffith's Hyperfine Splitting section for the Hydrogen atom, he concludes that the energy decrease for the singlet configuration is 3 times larger than the energy increase for the triplet configuration. While I can understand the mathematics of it, I can't see through it; I don't know how to explain it physically. Why would the singlet configuration have 3 times larger correction to the energy spectrum than a triplet configuration?

[Modified from Griffiths, Introduction to Quantum Mechanics (2nd edition)]

I think that the origin of the answer is in the more general question(not necessarily related to hydrogen atom and the hyperfine splitting):
In the case of spin-spin interaction between two fermions($s=\frac{1}{2}$) the operator $\vec{\sigma}_{1}\cdot\vec{\sigma}_{2}$ ($1$ and $2$ denoting each fermion respectively), why does the singlet state have an eigenvalue $-3$ times the eigenvalue of the triplet state?  Again, the mathematics are clear to me, I am just seeking to see it intuitively.
 A: If you want, you can reason that it's because there's three possibilities for the triplet but only one for the singlet. Quantum mechanics has to divide the triplet eigenvalue among three eigenvectors whereas the singlet eigenvalue belongs entirely to the singlet eigenvector.
Why I think things might be confusing
When you say "the mathematics are clear" I assume you mean the following standard argument: consider $\|\vec\sigma_1 + \vec\sigma_2\|^2 = \|\vec\sigma_1\|^2 + 2~ \vec \sigma_1\cdot\vec\sigma_2 + \|\vec\sigma_2\|^2$, we know that $\sigma^2$ eigenvalues go like $\hbar^2 \ell(\ell + 1)$ so that we should expect:$$\langle \vec \sigma_1\cdot\vec\sigma_2\rangle = \frac{\hbar^2}{2}\Big(s_{12}(s_{12} + 1) -s_1(s_1 + 1) - s_2(s_2 + 1)\Big),$$and using that both of the $s_{1,2}$ are $1/2$ gives
$$\langle \vec \sigma_1\cdot\vec\sigma_2\rangle = \hbar^2\Big(\frac{s_{12}(s_{12} + 1)}2 - \frac34\Big),$$
so we either see $1 - 3/4 = +1/4$ or $0 - 3/4 = -3/4$ for these two. Is that about right? Then yes, I can certainly see why this expression seems to come out of nowhere!
Now we're thinking with portals matrices
An alternative way to understand this is to look at the matrix representation; using the Pauli matrices and the Kronecker product, this dot product is just$$\begin{align}\sigma_x\otimes\sigma_x &+ \sigma_y\otimes\sigma_y + \sigma_z\otimes\sigma_z \\&= \begin{bmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{bmatrix}+\begin{bmatrix}0&0&0&-1\\0&0&1&0\\0&1&0&0\\-1&0&0&0\end{bmatrix}+\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{bmatrix}\\&=\begin{bmatrix}1&0&0&0\\0&-1&2&0\\0&2&-1&0\\0&0&0&1\end{bmatrix}.\end{align}$$where the superposition $a|\uparrow\uparrow\rangle + b|\uparrow\downarrow\rangle + c |\downarrow\uparrow\rangle + d|\downarrow\downarrow\rangle$ becomes the column vector $[a~~b~~c~~d]^T$. And probably you can diagonalize this in your sleep and see the four eigenvalues $1, 1, 1, -3,$ with the four eigenvectors $|\uparrow\uparrow\rangle,~~|\downarrow\downarrow\rangle,~~|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle,~~|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle.$ In addition I can give you a more intuitive feel for this middle term, which is that if $|\rightarrow\rangle = \sqrt{1/2}\big( |\uparrow\rangle + |\downarrow\rangle\big)$ in our coordinates then  $|\rightarrow \rightarrow\rangle,$  which is obviously "spin-1", must be a superposition of these "obviously spin-1" $|\uparrow\uparrow\rangle,~~|\downarrow\downarrow\rangle$ plus that middle term, so that middle term better be "obviously spin-1", too.
So the triplet/singlet degeneracy comes from these really low-level rules of quantum mechanics saying that when a particle is totally "spin-up" that means it's half "spin-left" and half "spin-right" and half "spin-forwards" and half "spin-backwards" because its total squared angular momentum $\hbar^2~\ell(\ell + 1)$ is larger than its angular momentum in its principal direction $\hbar~m.$ Aside from saying that a one-electron superposition of spin-up and spin-down gives us all the directions on the sphere, this also comes back to say that the four two-electron superpositions have 3 states where the spins align, and one state where the spins oppose. I can't get much more fundamental than that on these spin operators because they are sort of intrinsically hard to visualize concretely, that's kind of part of why QM is hard!
Tracery as sorcery
Okay now let's return to my intended answer: this antiparallel-spin "singlet state" is 3 times stronger in eigenvalue than the parallel-spin states. You can obviously see this for yourself above: but I said that that's because you have to spread the eigenvalue evenly among the three parallel-spin states; what the heck does that mean?
Well something like 80% of quantum mechanics is linear algebra in funny hats and so I am now going to invite you to remember back this concept of trace from linear algebra, where we both discovered that trace has linearity and cyclic permutivity laws; it therefore is the same for all similar matrices: and therefore, shockingly, the trace is the sum of the eigenvalues. Since an eigenvector of a Kronecker product $A\otimes B$ is going to be a vector $|\lambda_A,\lambda_B\rangle$ you're going to get all of the pairings, so actually the trace of $A \otimes B$ is going to be the trace of $A$ times the trace of $B$.
Well the Pauli matrices are all trace-free, and that's a simple result of the parity symmetry in our world, the assumption that things don't spin more in any dimension than they do in the opposite direction, so the eigenvalues all come as $\pm \lambda$ pairs. As a result all of these matrices $\sigma_i\otimes\sigma_i$ are trace-free too, and so their sum $\vec\sigma_1\cdot\vec\sigma_2$ must also be trace-free.
Now what is the trace? Here it's a sum $$\operatorname{Tr} X=\langle \uparrow\uparrow|X|\uparrow\uparrow\rangle + \langle \uparrow\downarrow|X|\uparrow\downarrow\rangle + \langle \downarrow\uparrow|X|\downarrow\uparrow\rangle + \langle \downarrow\downarrow|X|\downarrow\downarrow\rangle,$$ which if you divide by 4, is just the expected value of "I am going to flip 2 coins and choose an arbitrary spin configuration by that mechanism, and jam the system into that corresponding spin configuration."
The tracefree aspect of this $\vec\sigma_1\cdot\vec\sigma_2$ operator is really saying, "this modification to the Hamiltonian must on average disappear when we perform that choose-the-spin-configuration-at-random procedure," and you can kind of see that it has to do that, this is the sort of term which has no real directionality to it. But the fact that trace is basis-independent means that you can now carry this reasoning over to the singlet/triplet case: "if I choose one of these 4 configurations at random, I must recover on average only the energy of the level: however this perturbs that energy, it's got to average out."
And in that sense, the 3 levels which see an energy increase $+\epsilon$ must correspond with the one having an energy decrease $-3\epsilon$ so that if you choose a spin configuration at random, you get an average energy change of 0.
