# A tricky mutual information inequality

Let $$X_0, X_1, X_2$$ be three independently distributed bits, let $$B$$ be a random variable such that $$I(X_0:B)=0$$, $$I(X_1:B)=0$$, and $$I(X_2:B)=0$$, I need to prove that $$I(X_0, X_1, X_2:B)\leq 1$$

(where $$I(M:N)$$ is Shannon’s mutual information).

I can demonstrate that $$I(X_0, X_1, X_2:B)\leq 2$$, by using chain rule of mutual information $$I(X_0, X_1, X_2 : B)= I(X_0:B)+I(X_1,X_2:B|X_0) = H(X_1,X_2|X_0) – H(X_1,X_2|B,X_0) = 2 – H(X_1,X_2|B,X_0) \leq 2$$.

(where $$H(.)$$ is Shannon’s binary entropy).