Show that $\sqrt{\langle T(x), x \rangle}$ is a norm when $T$ is positive definite 
I'm having some trouble proving the second property of a norm, i.e. 
  $$\| x+y\| \leqslant \| x\| + \|y\|.$$
Let $T: R^n \to R^n$ be a linear operator and $\langle, \rangle: R^n\times R^n \to R$ be defined by 
$$\langle x,y\rangle=\sum_{i=1}^n x_iy_i \quad \forall \space x, y \in R^n.$$
Assume that $\langle T(x), x\rangle >0 \ \forall x \neq 0$ and 
  $\langle T(x), y \rangle=\langle T (y), x \rangle \ \forall x, y \in R^n$.
Prove that $f : R^n \to R^n $, defined by $f(x)= \sqrt{ \langle T (x), x\rangle} \quad \forall x \in R^n$ is a norm.

I already know that a solution to my problem is given by an auxiliary function $$g(t)= \langle T (x + ty), x + ty\rangle$$ through which I apply Cauchy-Schwartz inequality, but I proceed in another way, i.e.
$$
\begin{split}
f(x + y)^2
 &= \langle T (x+y),x + y\rangle \\
 &= \langle T (x), x + y\rangle + \langle T (y), x + y\rangle \\
 &= \langle T (x), x\rangle + \langle T (y), y\rangle
  + 2\langle T (x), y\rangle \\ ( f(x)+f(y) )^2 
 &= \langle T (x), x\rangle + \langle T (y), y\rangle
  + 2\langle T (x), x\rangle \langle T (y), y\rangle
\end{split}
$$
Thus, it remains to check if 
$$
\begin{split}
\langle T (x), y\rangle 
 & \leqslant \langle T (x), x\rangle \langle T (y), y\rangle \\
 &= \sum_{i=1}^n T(x_i)x_i \sum_{i=1}^n T(y_i)y_i \\
 &= \sum_{i=1}^n T(x_i)y_i \sum_{i=1}^n T(y_i)x_i \\
 &= 2\langle T (x), y\rangle.
\end{split}
$$
It follows that the inequality is proved. Is it correct?

I don't buy this:

$$ \sum_{i=1}^n T(x_i)x_i \sum_{i=1}^n T(y_i)y_i = \sum_{i=1}^n T(x_i)y_i \sum_{i=1}^n T(y_i)x_i $$

For example
$$ 45 =(1 \cdot 1 + 2 \cdot 2)(2 \cdot 2 + 3 \cdot 3) \ne (1 \cdot 2 + 2 \cdot 3)(2 \cdot 1 + 3 \cdot 2) = 64.$$
Instead, you have to prove Cauchy-Schwarz for the inner product $(x,y) \overset{\rm def} = \langle Tx, y \rangle$. Or, if you're allowed to, you can say that $(x,y)$ is an inner product and therefore Cauchy-Schwarz holds for it.
There are proofs of C-S that work for any inner product https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Proofs, so if you've seen one of those, it applies to $(x,y)$ and you don't need to prove it again.
Once you have C-S, the proof of the Triangle Inequality is a simple application:
\begin{align}
|(x,y)| &\le \|x\| \|y\| \\
\text{i.e. } |\langle Tx, y\rangle| &\le  \sqrt{\langle Tx, x \rangle} \cdot \sqrt{ \langle Ty, y \rangle}
\end{align}
Actually you made a slight error when you calculated $(f(x) + f(y))^2$, you should have this term: $$2f(x)f(y) = 2\sqrt{\langle Tx, x \rangle} \cdot \sqrt{\langle Ty, y \rangle}$$
instead of $2f(x)^2f(y)^2$.
