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Philomena
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$\begin{array}{l}a^{m+n} + b^{m+ n} \geq \dfrac{1}{2}(a^m + b^m)(a^n + b^n)\\ \Leftrightarrow 2a^m.a^n + 2b^m.b^n \geq a^m.a^n + a^m.b^n + b^m.a^n + b^m.b^n\\ \Leftrightarrow a^m.a^n + b^m.b^n \geq a^m.b^n + a^n.b^m\\ \Leftrightarrow a^m(a^n – b^n) + b^m(b^n – a^n) \geq 0\\ \Leftrightarrow (a^n – b^n)(a^m – b^m) \geq 0\quad (*)\\ +) \quad Khi\,\,a > b\\ \Rightarrow \begin{cases}a^n – b^n > 0\\a^m – b^m > 0\end{cases}\quad (m,n \in \Bbb N^*)\\ \Rightarrow (a^n – b^n)(a^m – b^m) > 0\\ \Rightarrow (*) \,\,đúng\\ +) \quad Khi\,\,a < b\\ \Rightarrow \begin{cases}a^n – b^n < 0\\a^m – b^m < 0\end{cases}\quad (m,n \in \Bbb N^*)\\ \Rightarrow (a^n – b^n)(a^m – b^m) > 0\\ \Rightarrow (*) \,\,đúng\\ +) \quad Khi\,\,a = b\\ \Rightarrow \begin{cases}a^n – b^n = 0\\a^m – b^m = 0\end{cases}\quad (m,n \in \Bbb N^*)\\ \Rightarrow (a^n – b^n)(a^m – b^m) = 0\\ \Rightarrow (*) \,\,đúng\\ \text{Vậy bất đẳng thức đã cho luôn đúng $\forall a, b > 0; \, m,n \in \Bbb N^*$} \end{array}$