Let $X$, $Y$ and $Z$ be positive definite Hermitian matrices (you ask about the real symmetric case, the Hermitian one includes it). Let the eigenvalues of $(X^*)^{-1/2} Y X^{-1/2}$ be $e^{\alpha_i}$, those of $(Y^*)^{-1/2} Z Y^{-1/2}$ be $e^{\beta_i}$ and those of $(X^*)^{-1/2} Z X^{-1/2}$ be $e^{\gamma_i}$. Set $A=Y^{1/2} X^{-1/2}$, $B=Z^{1/2} Y^{-1/2}$ and $C=Z^{1/2} X^{-1/2}$. The singular values of $A$ are then $e^{\alpha_i/2}$, and so forth.

Note that $AB=C$. By a result of Klyachko, there exist Hermitian matrices $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ such that $\mathfrak{a}+\mathfrak{b}=\mathfrak{c}$, the eigenvalues of $\mathfrak{a}$ are $\alpha_i/2$, those of $\mathfrak{b}$ are $\beta_i/2$ and those of $\mathfrak{c}$ are $\gamma_i/2$. This result can be thought of as saying that, although $\log A + \log B \neq \log C$, and although $\log A$, $\log B$ and $\log C$ are not Hermitian, we can find matrices which do have those properties and have the same eigenvalues.

The inequality you want to prove is that
$$\left( \sum \alpha_i^2 \right)^{1/2} + \left( \sum \beta_i^2 \right)^{1/2} \geq \left( \sum \gamma_i^2 \right)^{1/2}$$

But $\sum \alpha_i^2 = 4 \mathrm{Tr} \ \mathfrak{a}^* \mathfrak{a}$. So this turns into the (standard) fact that $\mathrm{Tr} \ \mathfrak{a}^* \mathfrak{a}$ is a positive definite norm on the Hermitian matrices.

I can't resist indulging in a little self promotion. This argument appears at the beginning of my paper Horn's Problem, Vinnikov Curves and the Hive Cone. There I consider the curve $\det(xX+yY+zZ)=0$ in $\mathbb{RP}^2$. This curve is hyperbolic and, by results of Helton and Vinnikov, all hyperbolic curves are of this form. It meets the three coordinate lines precisely at $-e^{\alpha_i}$, $-e^{\beta_i}$ and $-e^{\gamma_i}$. The point is to use results from the theory of hyperbolic curves to explain Horn's results on eigenvalues of matrix sums.

Analysis on symmetric cones, Oxford University Press (1994). – Denis Serre Jan 6 '11 at 12:34