Find constant of this inequality $\textbf{t}' \textbf{E}^b \textbf{t} \geq (\textbf{t}'\textbf{E} \textbf{t})^a$

Question

Let a vector $\textbf{t} \in \mathbb{R}^2$, $\textbf{E} \in \mathbb{R}^{2 \times 2}$ positive semidefinite matrix. Given that $$ \textbf{t}' \textbf{E}^b \textbf{t} \geq (\textbf{t}'\textbf{E} \textbf{t})^a,$$ $a = [0.5,1)$ and ' is the vector transpose. How can I obtain the real value of $b$?

Answer 1

For some matrices, there is no possible choice of $b$.

We can take, for example $$ E = \pmatrix{2\\&1/2} $$ This then becomes the inequality $$ 2^bx^2 + 2^{-b}y^2 \geq (2x^2 + y^2/2)^a $$ Setting $x=1,y=0$, we have the inequality $$ 2^b \geq 2^a \implies b \geq a $$ On the other hand, setting $x=0,y=1$, we have the inequality $$ 2^{-b} \geq 2^{-a} \implies b \leq a $$ which leaves us with $b = a$ as the only possibility. Then, as the comment below shows, even this is won't give us the desired inequlity.

So, no value of $b$ will work here for any $a$.