Ordering of inner products with constraints and matrix multiplication Suppose that $x^T y \ge z^T y$ for all vectors $x \in \mathbb{R}^d$.
If $H \in \mathbb{R}^{d \times d}$ is symmetric positive definite, then $x^T H y \ge z^T H y$.
Approach.
If $y_i \ge 0$ then we must have $x_i \ge z_i$. 
Otherwise we could construct $u$ to be the vector which is $z$ on all indices except $i$ but $u_i = x_i$.  This would achieve $u^T y \le z^T y$ which contradicts the hypothesis.
Likewise we can show that if $y_i \le 0$ then $x_i \le z_i$ and similar arguments can be made to show that we cannot have $y_i \ge 0$ and $x_i \le z_i$ (and vice-versa) as this would contradict the hypothesis.
Hence, $\sum_i (x_i - z_i)y_i > 0 $ is a sum of nonnegative terms.
Now is the bit that I am not sure about.
The matrix $H$ is positive definite so has positive eigenvalues $\lambda_i$.
In addition, $H$ is similar to a diagonal matrix $D_{ii} = (\lambda_i)$ whose nonzero entries are all positive.
Then $D_{ii}(x_i - z_i)y_i > 0$ for every $i$ so $\sum_i D_{ii}(x_i - z_i)y_i > 0$ which implies that $(x-z)^T D y > 0 $.
However, I don't want $D$ in this expression, I would like $H$.
Is there any way yo rectify the above argument to ensure the claim?
 A: The assumption 
$$
x^Ty \ge z^Ty \quad \forall x
$$
implies $y=0$. To see this, suppose $y\ne 0$. Then set $x := - t y$ for some positive $t>0$. Now, letting $t\to +\infty$ makes the left-hand side of the inequality arbitrarily small, while the right-hand side is bounded. A contradiction.
A: Note: Below $\langle\cdot,\cdot\rangle$ stands for the usual inner product of $\mathbb{R}^d.$ Moreover, the approach, that follows, doesn't begin on the same lines as yours. Let's see it.
The initial condition becomes $\langle x-z,y \rangle\geq 0,\ \forall x\in (F+z)$, where $F$ is a subspace of $\mathbb{R}^d$ (note from the answer of daw that this inequality cannot hold true for every $x\in \mathbb{R}^d$) and we want to prove that $\langle x-z,Hy\rangle,\ \forall x\in (F+z).$ We can take $x=w+z$ and then, the initial condition becomes 
$$\langle w,y\rangle\geq 0,\ \forall w\in F,\ \ \ (1)$$
while we now want to prove that
$$\langle w,Hy\rangle\geq 0,\ \forall w\in F.\ \ \ (2)$$
Now, we need extra conditions. One of them could be $H(F)=F$ (as the image of a linear map). In this case, for an arbitrary $w\in F,$ there exists $u\in F$ such that $w=H^{-1}u$ and we have that
$$\langle w,Hy\rangle=\langle H^{-1}u,Hy\rangle=\langle H^{t}H^{-1}u,y\rangle=\langle HH^{-1}u,y\rangle=\langle u,y\rangle\geq 0,$$
where the last inequality follows from $(1)$ and the second equality follows from the known property $\langle Ax_1,x_2\rangle=\langle x_1,A^{t}x_2\rangle.$ Note that $H$ is invertible, bacause all its eigenvalues are positive (a matrix is invertible if-f $0$ is not one of its eigenvalues). Since $w$ was arbitrary, $(2)$ is proved and we are done in this case. An other condition could be $Hy=y$. In this case
$$\langle w, Hy\rangle=\langle w,HH^{-1}y\rangle=\langle w,y\rangle\geq 0$$
again.  
