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Is this a valid use of the triangle inequality?

I know that I could not claim $\|x+y\| \leqslant \|x\| + \|y\| \stackrel{\text{X}}{\implies} \|x+y\|^{2} \leqslant \|x\|^{2} + \|y\|^{2} $ but does it hold in this way? \begin{align} \|x+y\|^{2} &= \|(x+z)+(y-z)\|^{2} \\ &\leqslant \|x+z\|^{2} + \|y-z\|^{2} \end{align}

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    No, you are using exactly the same property (which does not hold) – 2017-02-05
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    Indeed, following @Clement C. just set $z=0$ in your second statement. – 2017-02-05
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    If $x,y$ are orthogonal, then $||x+y||^2=||x||^2+||y||^2$ is the Pythagorean Theorem. – 2017-02-05
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    And if $y=2x$, $z=0$ for instance, then $\lVert x+y\rVert^2 = 9\lVert x\rVert^2$ while the RHS is $5\lVert x\rVert^2$ -- the inequality does not hold. – 2017-02-05

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Hint

Take $z=0$. What do you think?

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You have that $\|x+y\| \leq \|x\| + \|y\|,$ hence $$\|x+y\|^2 \leq \big(\|x\| + \|y\|\big)^2 = \|x\|^2+\|y\|^2+2\|x\|\|y\|. $$

Similarly,

$$\|(x+z)+(y-z)\|^2 \leq \|x+z\|^2+\|y-z\|^2+2\|x+z\|\|y-z\|.$$