Scalar product implying norm inequality

Question

Let $X$ be a Hilbert space. Fix $x_1, x_2, x_3\in X$. Suppose that $$\langle x_1,y\rangle_X\le \langle x_2,y\rangle_X+\langle x_3,y\rangle_X\qquad\mbox{for all $y\in X$}.\qquad (1)$$ Does
$$\|x_1\|_X\le \|x_2\|_X+\|x_3\|_X$$ follow from ($1$)?

Answer 1

Linearity of the inner product and rearranging gives:

$$\langle x_1 - x_2 - x_3, y\rangle \le 0, \quad\forall y \in X$$

In particular, for $y = x_1 - x_2 - x_3$ we get

$$0 \le \|x_1 - x_2 - x_3\|^2 = \langle x_1 - x_2 - x_3, x_1 - x_2 - x_3\rangle \le 0$$

Therefore $x_1 - x_2 - x_3 = 0$ so $x_1 = x_2 + x_3$.

Now apply the triangle inequality.

Answer 2

Hint: show first that the given inequality implies that $x_1=x_2+x_3$.

Answer 3

Hint:

Try letting $y=x_1$, $y=x_2$ and $y=x_3$ to get some expressions to work with.