Yes, there exists such a positive integer \(n\). It is well-known that the limit of RHS as \(n\to \infty\) is \(\max(a, b)\). We need to point out some positive integers \(a\) and \(b\) for which the following holds: \[\frac{\frac{a^2}{b}+\frac{b^2}{a}}{2} < \max(a, b).\]In fact take any \(a< b\). Then \(b=a+x\) with \(x>0\) and the above inequality becomes \[LHS-RHS=\frac 1 2 \frac{-x(a^2-ax+x^2)}{a(a+x)}<0.\]

Well yes, but now what is the maximal such $n$? That of course is the interesting part of the problem.

Tintarn wrote: Decide whether there exists a largest positive integer $n$ such that the inequality \[\frac{\frac{a^2}{b}+\frac{b^2}{a}}{2} \ge \sqrt[n]{\frac{a^n+b^n}{2}}\]holds for all positive real numbers $a$ and $b$. If such a largest positive integer $n$ exists, determine it. I think that answer is $n=9$. For $n=10$ we can choose $a=1,b=0.9$. Then $LHS<RHS$. It is known that $f(n)=\sqrt[n]{\frac{a^n+b^n}{2}}$ is monotonously increasing, so for $n>10$ inequality is also wrong. By the same reason it is enough to prove inequality for $n=9$. WLOG let $b=1$. Now inequality is equivalent with $(a^3+1)^9 \geqslant 2^8a^9(a^9+1)$. Now, for example, we can note that $$(a^3+1)^9 - 2^8a^9(a^9+1)=(a - 1)^4 (a + 1) (a^2 - a + 1) (a^2 + a + 1)^4 (a^{12} + 12 a^9 + 70 a^6 + 12 a^3 + 1) \geqslant 0$$and we are done. P.S. here is more clever proof of inequality $(a^3+1)^9 \geqslant 2^8a^9(a^9+1)$. Note that $$RHS=(a^3+1) \left (4^4 \cdot a^3 \cdot a^3 \cdot a^3 \cdot (a^6-a^3+1) \right ) \leqslant (a^3+1)(a^3+a^3+a^3+a^6-a^3+1)^4=(a^3+1)(a^6+2a^3+1)^4=(a^3+1)^9=LHS$$there we use AM-GM inequality.

Tintarn wrote: Decide whether there exists a largest positive integer $n$ such that the inequality \[\frac{\frac{a^2}{b}+\frac{b^2}{a}}{2} \ge \sqrt[n]{\frac{a^n+b^n}{2}}\]holds for all positive real numbers $a$ and $b$. If such a largest positive integer $n$ exists, determine it. See also here, here and here.

Strange for P6, the only obstruction is annoying calculations by hand... Equivalent to $(a^3 + b^3)^n\geq 2^{n-1}a^nb^n(a^n + b^n)$. Since the inequality is homogeneous and $a,b > 0$, we may assume without loss of generality that $b=1$ and so look for maximal $n$ with $(a^3+1)^n \geq 2^{n-1}a^n(a^n+1)$. Setting $a$ to be close to $1$ from below, say $a=\frac{9}{10}$, leads to $n\leq 9$. Finally, if $n=9$, then setting $x = a^3$, we want to show $(x+1)^9 \geq 256x^3(x^3+1)$, which by Horner's method (on the roots $1$ and $-1$) efficiently factorizes as $(x + 1)(x - 1)^4(x^4 + 12 x^3 + 70 x^2 + 12 x + 1)\geq 0$, done.