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d(x 0 , fx 0 ) = d(x n , fx n )d(x 0 , x 1 ) = d(x n , x n+1 ),d 0 = d nd 0 ≼ α n d 0 .As α = λ+µ1−δ < 1 (since λ + µ + δ < 1), the above inequality shows that d 0 = θ, i.e., d(x 0 , fx 0 ) = θ,so x 0 is a fixed point of f. Thus we assume that x n ≠ x m for all distinct n, m ∈ N.Again, as x n ⊑ x n+2 , we obtain from (I) and (4)d(x n , x n+2 ) = d(fx n−1 , fx n+1 )≼ λd(x n−1 , x n+1 ) + µd(x n−1 , fx n−1 ) + δd(x n+1 , fx n+1 )= λd(x n−1 , x n+1 ) + µd(x n−1 , x n ) + δd(x n+1 , x n+2 )≼ λ[d(x n−1 , x n ) + d(x n , x n+2 ) + d(x n+2 , x n+1 )] + µd(x n−1 , x n ) + δd(x n+1 , x n+2 )◭◭ ◭ ◮ ◮◮Go backFull ScreenCloseQuitso(5)= λ[d n−1 + d(x n , x n+2 ) + d n+1 ] + µd n−1 + δd n+1= (λ + µ)d n−1 + (λ + δ)d n+1 + λd(x n , x n+2 )≼ (λ + µ)α n−1 d 0 + (λ + δ)α n+1 d 0 + λd(x n , x n+2 )(λ + µ) + (λ + δ)α2d(x n , x n+2 ) ≼ α n−1 d 01 − λ≼ 2λ + µ + δ α n−1 d 0 ,1 − λwhere β = 2λ + µ + δ1 − λd(x n , x n+2 ) ≼ βα n−1 d 0 for all n ≥ 1,≥ 0.

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