Let $T \in \mathcal{L} (\mathcal{H})$ is positive operator (i.e. $\langle T x, x \rangle \ge 0$). Prove that $$\langle T^2 x, x \rangle \le \langle Tx, x \rangle^{\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^k}} \langle T^{2^k +1} x, x \rangle^\frac{1}{2^k}.$$ So I tried to use Cauchy-Schwarz inequality that is $$| \langle Tx, y \rangle| \le \langle Tx, x \rangle^\frac{1}{2} \langle Ty, y \rangle^\frac{1}{2}$$ and put $y = Tx$ which implies $$| \langle Tx, Tx \rangle| \le \langle Tx, x \rangle^\frac{1}{2} \langle T^2x, Tx \rangle^\frac{1}{2}.$$
Is it true that any positive operator in Hilbert space is self-adjoint and I can use that $T^* = T$ to get $$| \langle T^2x, x \rangle| \le \langle Tx, x \rangle^\frac{1}{2} \langle T^3x, x \rangle^\frac{1}{2},$$ which proves the base of induction?
It is often assumed (and practical) to assume that $T$ is selfadjoint when you say it's "positive". When $H$ is a complex Hilbert space, positivity in the sense of
implies $T^*=T$. In the real case it doesn't, but let us look at what the problem is.
The problem is, for instance, that if you only assume that $(1)$ holds, then $\langle Tx,x\rangle\geq0$ does not imply that $\langle T^2x,x\rangle\geq0$. Example: $$ \bigg\langle\begin{bmatrix}1&-2\\0&1 \end{bmatrix} \begin{bmatrix} a\\b\end{bmatrix} ,\begin{bmatrix} a\\b\end{bmatrix} \bigg\rangle\geq0 $$ for all $a,b$, while $$ \bigg\langle\begin{bmatrix}1&-2\\0&1 \end{bmatrix}^2 \begin{bmatrix} 1\\1\end{bmatrix} ,\begin{bmatrix} 1\\1\end{bmatrix} \bigg\rangle =\bigg\langle\begin{bmatrix}1&-4\\0&1 \end{bmatrix} \begin{bmatrix} 1\\1\end{bmatrix} ,\begin{bmatrix} 1\\1\end{bmatrix} \bigg\rangle =-2<0. $$