Assume that $\frac{1}{p}+\frac{1}{q}=1$ for two positive real numbers $p,q$.

For what kind of $C^*$ algebras $A$ the following hold:

$$\frac{ab+ba}{2}\leq \frac{a^p}{p}+\frac{b^q}{q}$$ $\forall a,b\in A^+$

As a second question we pose the trace version of this question:

Let $A$ be a $C^*$ algebra with a faitful trace is it true to say that for every two positive elements $a,b$ we have the following inequality? $$trace(ab)\leq \frac{trace(a^p)}{p} +\frac{trace(b^q)}{q}$$

Assume that $\frac{1}{p}+\frac{1}{q}=1$ for two positive real numbers $p,q$.

For what kind of $C^*$ algebras $A$ does the following hold:

$$\frac{ab+ba}{2}\leq \frac{a^p}{p}+\frac{b^q}{q}$$ $\forall a,b\in A^+$?

As a second question we pose the trace version of this question:

Let $A$ be a $C^*$ algebra with a faithful trace. Is it true to say that for every two positive elements $a$, $b$ we have the following inequality? $$\DeclareMathOperator\trace{trace}\trace(ab)\leq \frac{\trace(a^p)}{p} +\frac{\trace(b^q)}{q}$$

Assume that $\frac{1}{p}+\frac{1}{q}=1$ for two positive real numbers $p,q$.

For what kind of $C^*$ algebras $A$ the following hold:

$$\frac{ab+ba}{2}\leq \frac{a^p}{p}+\frac{b^q}{q}$$ $\forall a,b\in A^+$

As a second question we pose the trace version of this question:

Let $A$ be a $C^*$ algebra with a faitful trace is it true to say that for every two positive elements $a,b$ we have the following inequality? $$trace(ab)\leq \frac{trace(a^p)}{p} +\frac{trace(a^q)}{q}$$

Assume that $\frac{1}{p}+\frac{1}{q}=1$ for two positive real numbers $p,q$.

For what kind of $C^*$ algebras $A$ the following hold:

$$\frac{ab+ba}{2}\leq \frac{a^p}{p}+\frac{b^q}{q}$$ $\forall a,b\in A^+$

As a second question we pose the trace version of this question:

Let $A$ be a $C^*$ algebra with a faitful trace is it true to say that for every two positive elements $a,b$ we have the following inequality? $$trace(ab)\leq \frac{trace(a^p)}{p} +\frac{trace(b^q)}{q}$$