(∣z1∣+∣z2∣)∣z1∣z1+∣z2∣z2 Since ∣z1∣z1+∣z2∣z2≤∣z1∣z1+∣z2∣z2∣z1∣z1+∣z2∣z2≤∣z1∣∣z1∣+∣z2∣∣z2∣∣z1∣z1+∣z2∣z2≤2(∣z1∣+∣z2∣)(∣z1∣z1+∣z2∣z2)≤2(∣z1∣+∣z2∣)∴ statement I is correctFor Statement II :∣y−z∣a=∣z−x∣b=∣x−y∣c∣y−z∣2a2=∣z−x∣2b2=∣x−y∣2c2=λa2=λ(∣y−z∣2)=λ(y−z)(y−z)b2=λ(z−x)(z−x) and c2=λ(x−y)(x−y)y−za2+z−xb2+x−yc2=λ(y−z+z−x+x−y)=0Statement II is false