<s>[INST] Let $p_n$ denote the $n$th prime number. For $n \ge 3$, prove that $p_{n+3}^2 \lt p_np_{n+1}p_{n+2}$ Let $p_n$ denote the $n$th prime number.  For $n \ge 3$, prove that $p_{n+3}^2 \lt p_np_{n+1}p_{n+2}$.  
The book gives a hint as $p^2_{n+3} \lt 4p_{n+2}^2 \lt 8p_{n+1}p_{n+2}$, but I don't understand how to show that the hint is true either.
The only inequality relationship with prime squares that I know is Bonse's inequality but I can't seem to derive either the hint or the main problem from this.
 [/INST] For $n=3$, show it manually:
$(p_{n+3})^2=(p_{6})^2=13^2=169<385=5\cdot7\cdot11=(p_{3})(p_{4})(p_{5})=(p_{n})(p_{n+1})(p_{n+2})$

For $n=4$, show it manually:
$(p_{n+3})^2=(p_{7})^2=17^2=289<1001=7\cdot11\cdot13=(p_{4})(p_{5})(p_{6})=(p_{n})(p_{n+1})(p_{n+2})$

For $n\geq5$, use the fact that $p_{n+1}<2p_{n}$:
$(p_{n+3})^2<$
$(2p_{n+2})^2=$
$4(p_{n+2})(p_{n+2})<$
$4(2p_{n+1})(p_{n+2})=$
$8(p_{n+1})(p_{n+2})<$
$11(p_{n+1})(p_{n+2})=$
$(p_{5})(p_{n+1})(p_{n+2})\leq$
$(p_{n})(p_{n+1})(p_{n+2})$
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