Futher to my comments ...

For any $3$ consecutive primes (regardless of if they contain twin primes or not) $p_{n},p_{n+1},p_{n+2}$, it's true that $$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$ In fact, $\forall p_{n+k}\geq p_{n+2}$ we have $p_{n+k} > p_{n+1}$ and $p_{n+k} > p_{n}$, thus $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$ For a fixed $k$, we can show that $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$ This means that for large enough $n$'s we have $$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$ Applying AM-GM yields $$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}> \sqrt{p_{n+k}^2}=p_{n+k}$$ for large enough $n$'s, or $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n} \tag{2}$$ The range of $k$ depends on when $(1)$ gets less than $2$.

A few words about $(1)$, it is true because of this limit $$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ then $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}= \lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot \lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot \lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ again, for a fixed $k$!

Futher to my comments ...

For any $3$ consecutive primes (regardless of if they contain twin primes or not) $p_{n},p_{n+1},p_{n+2}$, it's true that $$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$ In fact, $\forall p_{n+k}\geq p_{n+2}$ we have $p_{n+k} > p_{n+1}$ and $p_{n+k} > p_{n}$, thus $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$ For a fixed $k$, we can show that $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$ This means that for large enough $n$'s we have $$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$ Applying AM-GM yields $$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}> \sqrt{p_{n+k}^2}=p_{n+k}$$ for large enough $n$'s, or $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n} \tag{2}$$ The range of the fixed $k$ depends on when $(1)$ gets less than $2$.

A few words about $(1)$, it is true because of this limit $$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ then $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}= \lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot \lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot \lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ again, for a fixed $k$!

Futher to my comments ...

For any $3$ consecutive primes (regardless of if they contain twin primes or not) $p_{n},p_{n+1},p_{n+2}$, it's true that $$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$ In fact, $\forall p_{n+k}\geq p_{n+2}$ we have $p_{n+k} > p_{n+1}$ and $p_{n+k} > p_{n}$, thus $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$ For a fixed $k$, we can show that $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$ This means that for large enough $n$'s we have $$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$ Applying AM-GM yields $$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}> \sqrt{p_{n+k}^2}=p_{n+k}$$ for large enough $n$'s, or $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n}$$ The range of $k$ depends on when $(1)$ gets less than $2$.

A few words about $(1)$, it is true because of this limit $$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ then $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}= \lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot \lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot \lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ again, for a fixed $k$!

Futher to my comments ...

For any $3$ consecutive primes (regardless of if they contain twin primes or not) $p_{n},p_{n+1},p_{n+2}$, it's true that $$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$ In fact, $\forall p_{n+k}\geq p_{n+2}$ we have $p_{n+k} > p_{n+1}$ and $p_{n+k} > p_{n}$, thus $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$ For a fixed $k$, we can show that $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$ This means that for large enough $n$'s we have $$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$ Applying AM-GM yields $$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}> \sqrt{p_{n+k}^2}=p_{n+k}$$ for large enough $n$'s, or $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n} \tag{2}$$ The range of $k$ depends on when $(1)$ gets less than $2$.

A few words about $(1)$, it is true because of this limit $$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ then $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}= \lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot \lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot \lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ again, for a fixed $k$!