From Ronald de Wolf:

Exercise 23.8 is wrong as stated. Counterexample: take X uniformly 
distributed over {0,1}, set Y=-X, and let alpha=1/2.  Then the random 
variable alphaX+(1-alpha)Y is always 0, so it has entropy 0.  On the 
other hand, H(X) and H(Y) are both equal to 1. So the lefthandside of 
the inequality is 0 while the righthandside is 1.  
<p>
I think it is the  probability distributions of X and Y that 
should be summed (with weights  alpha and 1-alpha), 
not their values themselves.  More precisely, assume 
X and Y are distributed over some set {1,...,n} (these two random 
variables may be dependent).  Define a new random variable Z, also 
distributed over {1,...,n}, with probability distribution 
Pr(Z=z)=alpha*Pr(X=z)+(1-alpha)Pr(Y=z).  Now prove the inequality 
H(Z)>=alpha*H(X)+(1-alpha)*H(Y).