Given a+b=1, prove the inequality a^2+b^2>ab.

Consider the linear model for a two-stage nested design with b nested in a as given below. yijk=\small \mu + \small \taui + \small \betaj(i) + \small \varepsilon(ij)k , for i=1,; j= ; k=1, assumption: \small \varepsilon(ij)k ~ iid n (0, \small \sigma2) ; \small \taui ~ iid n(0, \small \sigmat2 ); \tiny \sum_{j=1}^{b} \small \betaj(i) =0; \small \varepsilon(ij)k and \small \taui are independent. using only the given information, derive the least square estimator of \small \betaj(i) using the appropriate constraints (sum to zero constraints) and derive e(msb(a) ).