functional analysis - preserving norm map between real normed spaces

Suppose $X,Y$ are two real normed spaces,$T:X\to Y$ is the bijective map such that $||Tx+Ty+Tz|| = ||x+y+z||$ for any $x,y,z\in X$.Is $T$ a linear map?

Answer

By assumption
$$
\|T(x) + T(y) + T(-x-y)\| = 0,
$$
so
$$

T(x) + T(y) = -T(-x-y).
$$
In particular, with $x=y=0$ it follows $T(0)=0$, and with $y=0$,
$$
T(x) = -T(-x).
$$
Hence
$$
T(x) + T(y) = -T(-x-y) =T(x+y),
$$

and $T$ is additive.
Let me show that $T$ is continous. Let $x_n\to x$.
Then
$$
\|T(x) - T(x_n) \| = \|T(x) + T(-x_n)\|=\|x-x_n\|\to0.
$$
So $T$ is continuous and additive, hence it is linear, see Continuous and additive implies linear