Proving ineqalities for the similarity dimension 
a. Let $K$ be the attractor of the IFS $\{f_1,\dots f_n\}$ which satisfies SSC (i.e $f_i(K)\cap f_j(K)=\emptyset\forall i\neq j$) where for all $i, c_i$ such that $ 1\le i\le n, \space 0<c_i<1$ we have $$|f(y)-f(x)|\le c_i|y-x|$$Prove that $\dim_HK\le s$ s.t $\sum_i c_i^s=1$.
b. Let $H$ be attractor for this IFS such that $$|f(y)-f(x)|\ge c_i|y-x|$$ Prove that now $\dim_HH\ge s$

First using the given inequalities I thought that fixing $\alpha>0$ $$\mathscr{H}^\alpha(f_j(K))\le c_j^\alpha\mathscr{H}^\alpha(K)$$ and since $\mathscr{H}^\alpha(K)=\sum_{j=1}^\ell\mathscr{H}^\alpha(f_j(K))$ it follows that $\sum c_j^\alpha\ge 1$ and doing the same for (b), fixing $\beta$ we get $\sum_ic_j^\beta\le 1$. How in each part of the question we get the inverse inequality? 
EDIT: The question is defined on closed subset of $\mathbb{R}^d$ so I thought taking $\mu=\mathscr{L}^d$ but I cannot applicate it in (b)
 A: Your argument does not work if the $\alpha$-dimensional Hausdorff measure is $0$ or $\infty$, which it is for typical $\alpha$.
For part (a), if $(U_j)_{j}$ is any covering of $K$ with finitely many bounded sets, applying the IFS to it you get a new finite covering $(f_i(U_j))_{i,j}$ of $K$, and (using $|U|$ for the diameter of $U$)
$$
\sum_{i,j} |f_i(U_j)|^s \le \sum_{i,j} c_i^s |U_j|^s = \sum_j |U_j|^s.
$$
So if you repeatedly apply the IFS to any initial finite covering, the sums of the $s$-th powers of the diameters stay bounded, and the maximum diameter of the covering goes to $0$, which implies that the $s$-dimensional Hausdorff measure of $K$ is finite, so the Hausdorff dimension is at most $s$.
The standard proof for part (b) uses the mass distribution principle, where you use the IFS to define a self-similar measure and show that it satisfies an inequality like $\mu(B_r(x)) \le C r^s$, where $B_r(x)$ is the ball of radius $r$ centered at $x$. The details are a little more involved than the proof of (a). (However, that proof works under slightly weaker separation conditions, namely the open set condition. Maybe with the strong separation you are assuming there might be a shorter less technical proof.)
A good reference for all of this is Falconer's book on Fractal Geometry.
