Let #a=p/q# where #p# and #q# are positive integers.
#1ltp/q# therefore #qltp#. #p/qlt2# therefore #plt2q#. Therefore #qltplt2q#.
#a+1/a=p/q+q/p=(pp)/(qp)+(qq)/(pq)=(p^2+q^2)/(pq)=(p^2+2pq+q^2-2pq)/(pq)=(p+q)^2/(pq)-(2pq)/(pq)=(p+q)^2/(pq)-2#
#(q+q)^2/(qq)lt(p+q)^2/(pq)lt(2q+q)^2/(2qq)#*
#(2q)^2/q^2lt(p+q)^2/(pq)lt(3q)^2/(2q^2)#
#(4q^2)/q^2lt(p+q)^2/(pq)lt(9q^2)/(2q^2)#
#4lt(p+q)^2/(pq)lt9/2#
#4-2lt(p+q)^2/(pq)-2lt9/2-2#
#2lt(p+q)^2/(pq)-2lt5/2#
#2lta+1/alt5/2#
#5/2lt6/2#
#5/2lt3#
#2lta+1/alt3#
~~More advanced topics ahead~~
*This assumes that as #p# increases, #(p+q)^2/(pq)# increases. This can be verified intuitively, by looking at the graph of #y=(x+q)^2/(xq)# on #x in(q,2q)# for various positive values of #q#, or by the calculus process below.
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#del/(delp)[(p+q)^2/(pq)]=1/qdel/(delp)[(p+q)^2/p]=1/q(pdel/(delp)[(p+q)^2]-(p+q)^2del/(delp)[p])/p^2=1/q(p[2(p+q)]-(p+q)^2[1])/p^2=1/q(2p(p+q)-(p+q)^2)/p^2=((2p^2+2pq)-(p^2+2pq+q^2))/(p^2q)=(p^2-q^2)/(p^2q)#.
On #p in(q,2q)#:
Since #pgtqgt0#, #p^2gtq^2# thus #p^2-q^2gt0#.
Since #q>0#, #p^2qgt0#
Since #p^2-q^2gt0# and #p^2qgt0#, #(p^2-q^2)/(p^2q)gt0#
Since #del/(delp)[(p+q)^2/(pq)]=(p^2-q^2)/(p^2q)# and #(p^2-q^2)/(p^2q)gt0#, #del/(delp)[(p+q)^2/(pq)]gt0#
Therefore #(p+q)^2/(pq)# is increasing for constant #q# and #qltplt2q# because #del/(delp)[(p+q)^2/(pq)]# is positive.
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