To show that the inequality is true, you use mathematical induction

#1+1/sqrt2+...+1/sqrtn >=sqrt2(n-1)# for #n >1#

Step 1: Prove true for #n=2#

LHS=#1+1/sqrt2#

RHS=#sqrt2(2-1)=sqrt2#

Since #1+1/sqrt2 >sqrt2#, then #LHS > RHS#. Therefore, it is true for #n=2#

Step 2: Assume true for #n=k# where k is an integer and #k >1#

#1+1/sqrt2+...+1/sqrtk >=sqrt2(k-1)# --- (1)

Step 3: When #n=k+1#,

RTP: #1+1/sqrt2+...+1/sqrtk+1/sqrt(k+1) >=sqrt2(k+1-1)#

ie #0 >=sqrt2-(1+1/sqrt2+...+1/sqrtk+1/sqrt(k+1))#

RHS

=#sqrt2-(1+1/sqrt2+...+1/sqrtk+1/sqrt(k+1))#

#>=sqrt2-(sqrt2(k-1)+1/sqrt(k+1))# from (1) by assumption

=#sqrt2-sqrt2(k)+sqrt2-1/sqrt(k+1)#

=#2sqrt2-sqrt2(k)-1/sqrt(k+1)#

Since #k>1#, then #-1/sqrt(k+1)<0# and since #ksqrt2>=2sqrt2>0#, then #2sqrt2-ksqrt2<0# so #2sqrt2-sqrt2(k)-1/sqrt(k+1)=<0#

=LHS

Step 4: By proof of mathematical induction, this inequality is true for all integers #n# greater than #1#