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a)
x/y - (x+1)/(y+1) = [x(y+1)-(x+1)y]/ [y(y+1)] = (xy+x-xy-y)/[y(y+1)]
= (x-y)/[(y(y+1)]
x and y are natural numbers so if x/y>1 then x>y.
y>0 --> y(y+1)>0
x>y --> x-y >0
--> (x-y)/[y(y+1)] >0
<--> x/y - (x+1)/y+1)>0
<--> x/y > (x+1)/y+1)
(x+1)/(y+1)- 1 = [(x+1) – (y+1)] / (y+1) = (x+1-y-1)/(y+1) = (x-y)/(y+1) >0
-->(x+1)/(y+1)- 1 >0
-->(x+1)/(y+1) > 1
Finally, we have x/y > (x+1)/y+1) > 1
b) I will prove it by induction method.
Assume that Pk > √((k+1))
I will prove that Pk+2 > √((k+1+2) ) = √((k+3))
so Pk * (k+2)/(k+1) > √(k+3)
I will prove that Pk * (k+2)/(k+1) > √((k+1)) *(x+2)/(x+1) > √(k+3)
√((k+1)) *(x+2)/(x+1) > √(k+3)
<-> (k+2)/(k+1)> √((k+3)/(k+1))
<-> ((k+2)/(k+1))^2> (k+3)/(k+1)
<-> (k^2+4k+4)/(k^2+2k+1)> (k+3)/(k+1)
<-> 1+ (2k+3)/(k^2+2k+1)> 1+ 2/(k+1)
<-> (2k+3)/(k^2+2k+1)> 2/(k+1)
<-> (2k+3)/(k^2+2k+1)- 2/(k+1)>0
<-> ((2k+3)(k+1)-2(k^2+2k+1))/(k^2+2k+1)(k+1) >0
<-> (2k^2+5k+3-2k^2-4k-2)/(k^2+2k+1)(k+1) >0
<->(k+1)/(k^2+2k+1)(k+1) >0 ( of course because k>0)
So we have: √((k+1)) *(x+2)/(x+1) > √(k+3) and we already assumed that Pk > √((k+1)) so that Pk * (k+2)/(k+1) > √((k+1)) *(x+2)/(x+1) > √(k+3)  Pk * (k+2)/(k+1) > √(k+3). As a result, Pk+2 > √((k+3)).
Now we have, if Pk > √((k+1)) then Pk+2 > √((k+3)).
Substitute to k =2 or 4 we can say that is true so the inequality has been solved.
c) In this case, I still use induction method, but I will prove that if Pk > √((3k/2+1))
Assume that Pk > √((3k/2+1))
So I will prove that Pk+2 = Pk * (k+2)/(k+1) > √((3k/2+1)) * (k+2)/(k+1) > √(((3(k+2))/2+1)) = √((3k/2+4))
I will start proving √((3k/2+1)) * (k+2)/(k+1) > √((3k/2+4))
→(k+2)/(k+1)>√((3k/2+4)/(3k/2+1))
<-> (k^2+4k+4)/(k^2+2k+1)>(3k/2+4)/(3k/2+1)=1+ (2k+3)/(k^2+2k+1)>1+3/(3k/2+1)
<->(2k+3)/(k^2+2k+1)> 3/(3k/2+1)
<->(2k+3)/(k^2+2k+1)- 3/(3k/2+1)>0
<->((2k+3)(3k/2+1)-3(k^2+2k+1))/((k^2+2k+1)(3k/2+1))>0
<->(3k^2 4.5k+2k+3-〖3k〗^2-6k-3)/((k^2+2k+1)(3k/2+1))>0
<->0.5k/((k^2+2k+1)(3k/2+1))>0 (Of course with all k>0)
So we have √((3k/2+1)) * (k+2)/(k+1) > √((3k/2+4)) <-> Pk+2 = Pk * (k+2)/(k+1) > √((3k/2+1)) * (k+2)/(k+1) > √((3k/2+4))  Pk+2 > √((3k/2+4))

We have conclusion that if Pk > √((3k/2+1)) then Pk+2 > √((3k/2+4))
Substitute to k = 3 or 4 or 5 we can see that is true. Apply to the case k =100 we have P100 > √(((3*100)/2+1))=√151>√144=12 so Q = P100>12