Let me expand slightly on the comments I made above, and give the most general solution.
Clearly the inequality $\frac{ab + ba}{2} \leq \frac{a^2}{2} + \frac{b^2}{2}$ holds for all positive elements $a,b$ in a $C^\ast$-algebra since $a-b$ is self-adjoint and therefore $0\leq (a-b)^2 = a^2 + b^2 - ab - ba$.
Also (just to get it out of the way before we get technical), the paper [Farenick, Douglas R., Manjegani, S. Mahmoud, Young's inequality in operator algebras. J. Ramanujan Math. Soc. 20 (2005), no. 2, 107–124] shows that for all $p,q >1$ with $\tfrac{1}{p} + \tfrac{1}{q} = 1$ all $C^\ast$-algebras $A$ with a faithful tracial state $\tau$, the inequality $\tau(ab) \leq \frac{\tau(a^p)}{p} + \frac{\tau(b^q)}{q}$ holds for all $a,b\in A_+$.
Here comes the technical part:
Proposition: Let $A$ be a $C^\ast$-algebra and let $p,q>1$ such that $\tfrac{1}{p} + \tfrac{1}{q} = 1$ and $p\neq q$. Then the inequality $\frac{ab+ba}{2} \leq \frac{a^p}{p} + \frac{b^q}{q}$ holds for all $a,b\in A_+$ if and only if $A$ is commutative.
Proof: If $A$ is commutative, then $A \cong C_0(X)$ for some locally compact Hausdorff space $X$, and the inequality in question holds by Young's inequality since the order relation $\leq$ in $C_0(X)$ is the pointwise $\leq$ of complex numbers.
Conversely, suppose $A$ is non-commutative. A classical result (which I think might go back to Kadison (I have added a proof below)) shows that $A$ contains a $C^\ast$-subalgebra which surjects onto $M_2(\mathbb C)$. As positive elements in quotients of $C^\ast$-algebras lift to positive elements (by functional calculus), it suffices to witness the failure of our desired inequality in $M_2(\mathbb C)$.
Assume without loss of generality that $p>2$. Let $a(\epsilon) = \left(\begin{array}{cc} \epsilon & 0 \\ 0 & 0\end{array}\right)$ for $\epsilon>0$ and $b = \left(\begin{array}{cc} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right)$. We will show that $a(\epsilon)$ and $b$ fail the inequality for sufficiently small $\epsilon$. Note that $a(\epsilon)^p = a(\epsilon^p)$ and that $b^q = b$ (since $b$ is a projection). So we should verify that the
\begin{equation}
\frac{a(\epsilon^p)}{p} + \frac{b}{q} - \tfrac{1}{2}(a(\epsilon)b + ba(\epsilon)) = \left( \begin{array}{cc} \tfrac{\epsilon^p}{p} + \tfrac{1}{2q} - \tfrac{\epsilon}{2} & \tfrac{1}{2q} - \tfrac{\epsilon}{4} \\ \tfrac{1}{2q} - \tfrac{\epsilon}{4} & \tfrac{1}{2q} \end{array} \right)
\end{equation}
is not positive semidefinite for small $\epsilon$. We do this by showing that it has a negative determinant. In fact, the determinant of the above matrix is
\begin{equation}
\frac{\epsilon^p}{2pq} + \frac{1}{4q^2} -\frac{\epsilon}{4q} - \frac{1}{4q^2} - \frac{\epsilon^2}{16} + \frac{\epsilon}{4q} = \epsilon^2 (\frac{\epsilon^{p-2}}{2pq} - \frac{1}{16})
\end{equation}
and this is clearly negative for sufficiently small $\epsilon >0$. QED.
ADDON: Upon request I will add the following which I think is due to Kadison (the driving force is Kadison transitivity). I have added most details, but some minor elementary things are left for the reader.
Proposition: Let $A$ be a $C^\ast$-algebra. The following are equivalent:
(i) $A$ is non-commutative;
(ii) there exists an irreducible representation $\pi \colon A \to \mathcal B(H)$ where $\mathrm{dim}(H) \geq 2$;
(iii) $A$ contains a (non-zero) nilpotent element;
(iv) there exists a $C^\ast$-subalgebra $B \subseteq A$ which contains a two-sided closed ideal $J$ such that $B/J \cong M_2(\mathbb C)$.
Proof:
(iv) $\Rightarrow$ (i) is obvious, (i) $\Rightarrow$ (ii) is standard (irreducible representations separate points, so if every irreducible representation is 1-dimensional, then all commutators $ab-ba$ in $A$ vanish).
(ii) $\Rightarrow$ (iii): Fix an irreducible representation $\pi \colon A \to \mathcal B(H)$ where $\mathrm{dim}(H) \geq 2$. Let $\xi, \eta \in H$ be orthogonal unit vectors. By Kadison transitivity there are elements $a,b\in A$ such that
\begin{equation}
\pi(a) \xi = \tfrac{1}{2}\xi , \quad \pi(a) \eta = \eta, \quad \pi(b) \xi = \eta \quad \pi(b) \eta = \xi.
\end{equation}
Replacing $a$ with $|a|$ we may assume $a\geq 0$. Let $f,g\in C_0((0,\|a\|])$ such that $fg = 0$ and $f(1/2) = 1$ and $g(1) = 1$. I claim that $x= f(a) b g(a) \in A$ is nilpotent of order 2. In fact, clearly $x^2 = 0$ since $g(a) f(a) = 0$ (by functional calculus), so we should show that $x\neq 0$. Since $f(\pi(a)) \xi = \xi$ and $g(\pi(a)) \eta = \eta$ we have
\begin{equation}
\pi(x) \eta = f(\pi(a)) \pi(b) g(\pi(a)) \eta = f(\pi(a)) \pi(b) \eta = f(\pi(a)) \xi = \xi
\end{equation}
so $x\neq 0$.
(iii) $\Rightarrow$ (iv): Represent $A\subseteq \mathcal B(H)$ faithfully (not the same $H$ as above), and let $x\in A$ be a nilpotent element of order 2. We may assume $\|x\| =1$. Let $x = u|x|$ be the polar decomposition of $x$ in $\mathcal B(H)$. Then $u$ is a partial isometry with orthogonal range and source projections, and thus $C^\ast(u) \cong M_2(\mathbb C)$. Let $y= |x| + |x^\ast|$ (which is the sum of two orthogonal positive elements, so $\|y\|=1$). Then $yu = uy$ and $u^\ast y =y u^\ast$ and thus there is a unique $\ast$-homomorphism $\psi \colon C_0((0,1]) \otimes M_2(\mathbb C) \to \mathcal B(H)$ such that $\psi(f\otimes e_{1,2}) = f(y) u$. The image of $\psi$ takes values in $A$ since $y^k u = y^{k-1} x \in A$, so $f(y)u \in A$ for any polynomial $f$ with trivial constant term. Let $B = \psi(C_0((0,1])\otimes M_2(\mathbb C))$ (which is equal to $C^\ast(x)$ but this is irrelevant) and $J = \psi(C_0(0,1) \otimes M_2(\mathbb C))$. Then $B/J \cong M_2(\mathbb C)$.
