Paper 4, Section I, GNumber TheoryPart II, 2017Show that, for x⩾2x \geqslant 2x⩾2 a real number,∏p⩽xp is prime (1−1p)−1>logx\prod_{\substack{p \leqslant x \\ p \text { is prime }}}\left(1-\frac{1}{p}\right)^{-1}>\log xp⩽xp is prime ∏(1−p1)−1>logxHence prove that∑p⩽x,p is prime 1p>loglogx+c,\sum_{\substack{p \leqslant x, p \text { is prime }}} \frac{1}{p}>\log \log x+c,p⩽x,p is prime ∑p1>loglogx+c,where ccc is a constant you should make explicit.