In $\mathbb{R^2}$, $\langle (x_1,y_1), (x_2,y_2) \rangle=x_1x_2-\alpha(x_1y_2+x_2y_1)+y_1y_2$ is an inner product 
In $\mathbb{R^2}$, $\langle (x_1,y_1), (x_2,y_2)
 \rangle=x_1x_2-\alpha(x_1y_2+x_2y_1)+y_1y_2$ is an inner product
(a) $\forall \alpha\in \mathbb R$ 
(b) iff $\alpha=0$
(c)iff $\alpha <1$
(d) iff $ | \alpha | <1$

Using the first axiom of IPS,
 $\langle (x_1,y_1), (x_1,y_1)\rangle \ge 0$ $\implies$ $\frac{x_1^2+y_1^2}{2 x_1 y_1}\ge \alpha$ applying the A.M-G.M Inequality then I got $\alpha<1$. I know, when $\alpha =0$ also it is true. So, (a), (b) are false. How to judge the answer?
 A: The answer is for all $|\alpha|<1.$ Consider $$A=\begin{bmatrix}1&-\alpha\\-\alpha&1\end{bmatrix}.$$ Then $\langle x,y\rangle$ as you have defined it is equal to $\langle Ax,y\rangle_{E},$ using the standard (i.e., Euclidean) inner product. It is an easy exercise to show that $\langle A\cdot,\cdot\rangle_{E}$ defines an inner product if and only if $A$ is positive definite, which is true if and only if $\mathrm{det}(A)>0,$ since $A$ has the form above (and in particular, the diagonal entries are positive, so if this holds, the leading principal minors are all positive, which is sufficient for positive definiteness). Since $\mathrm{det}(A)=1-|\alpha|^{2},$ this is positive if and only if $|\alpha|<1.$
We show that $\langle Ax,y\rangle_{E}$ defines an inner product if and only if $A$ is positive definite. If this is an inner product $\langle\cdot,\cdot\rangle_{A}$, then when $x=y,$ we see that $x^{*}Ax=\langle x,x\rangle_{A}\geq0,$ and equals $0$ if and only if $x=0$ by properties of the inner product. But this proves that $A$ is positive definite.
Conversely, if $A$ is positive definite, letting $\langle x,y\rangle_{A}=\langle Ax,y\rangle_{E}$, we see that $\langle x,x\rangle_{A}\geq0$ for all $x\in\mathbb{R}^{n}$, and is positive whenever $x\neq0,$ by definition of positive definiteness. Since $\langle x,y\rangle_{A}=y^{*}Ax,$ $\langle y,x\rangle_{A}=x^{*}Ay=(y^{*}Ax)^{*}=\overline{\langle x,y\rangle_{A}},$ using the fact that $A=A^{*}.$ Finally, $\langle ax+by,z\rangle_{A}=z^{*}A(ax+by),$ and by linearity of matrix multiplication, this is $az^{*}Ax+bz^{*}Ay=a\langle x,z\rangle_{A}+b\langle y,z\rangle_{A}.$ This completes the proof.
