Apply the Frobenius map to minimal polynomials to see that $$ [K(\alpha^p+\beta^p):K(\alpha^p+\beta^p,\beta^p)]\leq [K(\alpha+\beta):K(\alpha+\beta,\beta)] $$ and $$ [K(\alpha^p+\beta^p):K] \leq [K(\alpha+\beta):K] $$
Let $\overline{K}$ be a fixed algebraic closure. Let $m_1$ be the minimal polynomial of $\beta^p$ over $K(\alpha^p+\beta^p)$ and let $m_2$ be the minimal polynomial of $\beta$ over $K(\alpha+\beta)$. If I can show that $m_1 \mid m_2$ in $\overline{K}[x]$ then this would be enough. But here I'm stuck, because I'm not sure how to use the Frobenius map, i.e. $(a+b)^p=a^p+b^p$, to show this. Similarly for the second inequality. Reference material is posted below.
Let $F:\bar{K}\to\bar{K},x\mapsto x^p$ denote the Frobenius map. Note that $F(K(\alpha+\beta))\subseteq K(\alpha^p+\beta^p)$. Consider the polynomial $F(m_2)$ ($F$ applied to the coefficients). Then $F(m_2)\in K(\alpha^p+\beta^p)[x]$ and $F(m_2)(\beta^p)=F(m_2)(F(\beta))=F(m_2(\beta))=0$. Thus $m_1\mid F(m_2)$ and therefore \begin{align*} [K(\alpha^p+\beta^p,\beta^p):K(\alpha^p+\beta^p)]=\deg m_1&\leq\deg F(m_2)\\&=\deg m_2=[K(\alpha+\beta,\beta):K(\alpha+\beta)] \end{align*} The second inequality follows from $K(\alpha^p+\beta^p)\subseteq K(\alpha+\beta)$.