Question:

**Step 1:** Analyze the given inequality. The given inequality is $\dfrac{p^2 + q^2}{2} \geq \sqrt{p^2 q^2}$. Since $p, q > 0$, we have $\sqrt{p^2 q^2} = pq$. Thus, the inequality simplifies to $p^2 + q^2 \geq 2pq$. This is equivalent to $(p-q)^2 \geq 0$, which is always true for any real numbers $p$ and $q$. This inequality holds for all $p,q > 0$. **Step 2:** Express $p+q$ in terms of $p^2+q^2$ and $pq$. To find the maximum value of $p+q$, consider the square of $p+q$: $$(p+q)^2 = p^2+q^2+2pq$$ **Step 3:** Determine the maximum value under a specific condition. For $p+q$ to have a finite maximum value, $p^2+q^2$ must be bounded. Let us consider the condition $p^2+q^2=1$. Under this condition, the inequality $p^2+q^2 \geq 2pq$ (from Step 1) becomes $1 \geq 2pq$, which implies $pq \leq \dfrac{1}{2}$. Substitute $p^2+q^2=1$ into the expression for $(p+q)^2$: $$(p+q)^2 = 1+2pq$$ To maximize $p+q$ (and thus $(p+q)^2$ since $p+q>0$), we must maximize $pq$. The maximum value of $pq$ is $\dfrac{1}{2}$, which occurs when $p=q$. If $p=q$ and $p^2+q^2=1$, then $2p^2=1$, so $p^2=\dfrac{1}{2}$, which implies $p=q=\dfrac{1}{\sqrt{2}}$. **Step 4:** Calculate the maximum value of $p+q$. Substitute the maximum value of $pq = \dfrac{1}{2}$ into the expression for $(p+q)^2$: $$(p+q)^2 = 1+2\left(\dfrac{1}{2}\right) = 1+1 = 2$$ Since $p,q > 0$, $p+q > 0$. Taking the positive square root, we find the maximum value: $$p+q = \sqrt{2}$$