(1)
f(x)=ln(1+x) -x(1+ λx)/(1+x)
f(0) = 0
ie To find min λ st f'(x) ≤ 0
f'(x) = 1/(1+x) - [(1+x)(1+2λx)-x(1+λx) ]/(1+x)^2
={1+x -[(1+x)(1+2λx)-x(1+λx) ]}/(1+x)^2
={1+x -[1+(2λ+1)x+2λx^2-x-λx^2) ]}/(1+x)^2
= ( -λx^2 -(2λ-1)x )/(1+x)^2
= -(λx^2 +(2λ-1)x )/(1+x)^2
2λ-1=0
min λ =1/2 st. f'(x) ≤0
f(x)=ln(1+x) -x(1+ λx)/(1+x) ≤f(0) =0
(2)
By MI
n=1
a2 -a1 +1/4
=1+1/2-1 +1/4
=3/4>ln2
p(1) is true
Assume p(k) is true
a(2k)-ak +1/(4k) > ln2
for n=k+1
LS =a(2k+2)-a(k+1) - 1/[4(k+1)]
= 1/(k+2)+1/(k+3)+...+1/(2k+2) +1/[4(k+1)]
=[1/(k+1)+1/(k+2)+...+1/(2k) +1/(4k) ] +1/(2k+1)+1/(2k+2) - 1/(k+1)- 1/(4k) + 1/[4(k+1)]
> ln2 + 1/(2k+1)+1/(2k+2) - 1/(k+1)- 1/(4k) + 1/[4(k+1)]
= ln2 + 1/(2k+1)-1/[4(k+1)] - 1/(4k)
consider
1/(2k+1)-1/[4(k+1)] - 1/(4k)
= [16k(k+1)-4k(2k+1)-4(2k+1)(k+1)] /[16k(2k+1)(k+1)]
= (6k^2+9k-2)/[16k(2k+1)(k+1)] >0
=>p(k+1) is true
By principle of MI, it is true for all +ve integer n