Consider a graph $G$ with max degree $\Delta_G$, min degree $\delta_G$ and average degree $d_G$. Is it possible to obtain a subgraph of $G$, say $G'$, such that  $\Delta_{G'} = c_1d_{G}$, $\delta_{G'} = c_2d_{G}$ , where $c_1,c_2$ are constantsEDIT 1: Also the size of the resultant subgraph $G'$ must be constant times the size of original graph $G$.My attempt : Let $X$ be the random variable which denotes the degree of any randomly chosen vertex. Therefore, $d_G = E[X]$. Using markovs's inequality, we have$$P(X > 3E[X]) < \frac{1}{3} $$ and $$P(X < \frac{E[X]}{3}) = P(\frac{1}{X} > \frac{3}{E[X]}) < \frac{E[\frac{1}{X}]*E[X]}{3} < \frac{1}{3}$$Using union bound, we have$$P(\frac{E[X]}{3} < X < 3E[X]) > 1/3$$.Using the above equation, we can show that there exist $|V|/3$ vertices in $G$ whose degree is between $d_G/3$ and $3d_G$.Does the graph obtained by the above $|V|/3$ vertices solve the problem? If not, can we modify the technique to obtain a solution?Thanks in advance!
Subgraph of G whose maximum degree and minimum degree are of the same order
graph theory;pr.probability