Let $A$ and $B$ be two (non commuting) self-adjoint bounded operator acting on a Hilbert space and let $p,q>1$ such that $\frac1p+\frac1q=1$


Do we have a Young-type inequality such as $ \frac12|AB+BA| \leq \frac{|A|^p}{p}+ \frac{|B|^q}{q}$ in the sense of quadratic form ? It is of course obvious for $p=q=2$.



Young-type inequality for bounded operator
As a general criterion, these kind of inequalities hold at the level of singular values (which implies a version of the inequality where one of the sides is conjugated by a unitary/partial isometry), but they tend to fail at the plain operator level.


Here is a counterexample for $p=3$, $q=3/2$. Take
$$
A=\begin{bmatrix}1&0\\0&3\end{bmatrix},\qquad\qquad B=\begin{bmatrix}1&1\\1&1\end{bmatrix}.
$$
Then
$$
\frac{A^3}3+\frac{2B^{3/2}}3=\frac13\,\Big(\begin{bmatrix}1&0\\0&27\end{bmatrix}
+\begin{bmatrix}2\sqrt2&2\sqrt2\\2\sqrt2&2\sqrt2\end{bmatrix}\Big) =\frac13\,\begin{bmatrix}1+2\sqrt2&2\sqrt2\\2\sqrt2&27+2\sqrt2\end{bmatrix}.
$$
We have
$$
AB+BA=\begin{bmatrix}2&21\\21&40\end{bmatrix},
$$
so
$$
(AB+BA)^2=\begin{bmatrix} 20&32\\32&52\end{bmatrix}
$$
and then [Wolfram Alpha](https://www.wolframalpha.com/input?i=simplify+MatrixPower%5B%7B%7B20%2C32%7D%2C%7B32%2C52%7D%7D%2C1%2F2%5D) tells us that
$$
|AB+BA|=\begin{bmatrix} 20&32\\32&52\end{bmatrix}^{1/2}=\frac2{\sqrt5}\,\begin{bmatrix}3& 4\\ 4& 7
\end{bmatrix}.
$$


Then, [using Wolfram Alpha](https://www.wolframalpha.com/input?i=%281%2F3%29%7B%7B1%2C0%7D%2C%7B0%2C3%5E3%7D%7D%2B%282sqrt%282%29%2F3%29%7B%7B1%2C1%7D%2C%7B1%2C1%7D%7D-%281%2F2%29%7B%7B6%2Fsqrt%285%29%2C+8%2Fsqrt%285%29%7D%2C+%7B8%2Fsqrt%285%29%2C+14%2Fsqrt%285%29%7D%7D) again, we see that
$$
\frac{A^3}3+\frac{2B^{3/2}}3-\frac12\,|AB+BA|
$$
is not positive, as it has a negative eigenvalue.


