$$
H(p,q) \geq H(p) \xrightarrow[]{} \sum_{x}^{}-p_{x}\log(q_{x}) \geq \sum_{x}^{}-p_{x}\log(p_{x}) \\ \xrightarrow[]{-\log(x)=\log(\frac{1}{x})} \sum_{x}^{}p_{x}\log(\frac{1}{q_{x}}) \geq \sum_{x}^{}p_{x}\log(\frac{1}{p_{x}})
\\ \xrightarrow[]{} \sum_{x}^{}p_{x}\log(\frac{1}{p_{x}}) - \sum_{x}^{}p_{x}\log(\frac{1}{q_{x}})  \leq 0
\\ \xrightarrow[]{} \sum_{x}^{}p_{x}[\log(\frac{1}{p_{x}})- \log(\frac{1}{q_{x}})] \leq 0
\\ \xrightarrow[]{\log(x) - \log(y) = \log(\frac{x}{y})} \sum_{x}^{}p_{x}\log(\frac{q_{x}}{p_{x}}) \leq 0  
\\ \xrightarrow[\sum_{x}^{}p_{x} = 1, So ItIsLike {\color{Red} \alpha} In Concave Definition]{{\log is a {\color{Red} Concave Function}}} \log[\sum_{x_{p_{x}\neq 0}}^{}p_{x}(\frac{q_{x}}{p_{x}})] \leq 0 
\\ \xrightarrow[]{} \log[\sum_{x_{p_{x}\neq 0}}^{}\not{p_{x}}\frac{q_{x}}{\not{p_{x}}}] \leq 0
\xrightarrow[]{} \log(\sum_{x_{p_{x}\neq 0}}^{}q_{x}) \leq 0 
\\ \xrightarrow[]{} \log(\sum_{x_{p_{x}\neq 0}}^{}q_{x}) \leq \log(\sum_{x}^{}q_{x}) =  \log(1) = 0
\\ \xrightarrow[]{} Proved $$