Given $f(\mathbf{x})=\sum_{i=1}^{n}x_i\ln(x_i)$, I want to prove that the function:
$$G(\mathbf{u},\mathbf{v})=f(\mathbf{u})-f(\mathbf{v})-(\mathbf{u}-\mathbf{v})^{T}\nabla f(\mathbf{v})$$
is non-negative for $\mathbf{u}$ and $\mathbf{v}$ strictly positive and identically zero iff $\mathbf{u}=\mathbf{v}$. My attempt so far for the inequality consisted of finding $\nabla f(\mathbf{v})=\ln(\mathbf{v})+\mathbf{1}$ and doing some algebra:
\begin{align} f(\mathbf{u})-f(\mathbf{v})-(\mathbf{u}-\mathbf{v})^{T}\nabla f(\mathbf{v})&\geq0\\ f(\mathbf{u})-f(\mathbf{v})&\geq(\mathbf{u}-\mathbf{v})^{T}\nabla f(\mathbf{v})\\ \sum_{i=1}^{n}u_i\ln(u_i)-\sum_{i=1}^{n}v_i\ln(v_i)&\geq(\mathbf{u}-\mathbf{v})^{T}(\ln(\mathbf{v})+\mathbf{1})\\ \ln\left(\prod_{i=1}^{n}\frac{u_{i}^{u_i}}{v_{i}^{v_i}}\right)&\geq\sum_{i=1}^{n}u_i\ln(v_i)-v_i\ln(v_i)+u_i-v_i\\ \ln\left(\prod_{i=1}^{n}\frac{u_{i}^{u_i}}{v_{i}^{v_i}}\right)&\geq\ln\left(\prod_{i=1}^{n}\frac{v_{i}^{u_i}}{v_{i}^{v_i}}\right)+\sum_{i=1}^{n}u_i-v_i\\ \ln\left(\prod_{i=1}^{n}\left(\frac{u_{i}}{v_{i}}\right)^{u_i}\right)&\geq\sum_{i=1}^{n}u_i-v_i\\ \prod_{i=1}^{n}\left(\frac{u_{i}}{v_{i}}\right)^{u_i}&\geq\prod_{i=1}^{n}\frac{e^{u_i}}{e^{v_i}} \end{align}
And here I got stuck...
Is what I did so far correct? Is this inequality somewhat obvious or trivial? I know that the original inequality can be easily proven by taking the convex properties of $f$ into account and using Mean-Value Theorem. However, I would like to see if I could reach the same conclusion by the path I was taking.
If your inequality holds for all $u_i,v_i$, then it also holds without the product. Thus, you want to prove that $$ \left(\frac uv\right)^u > e^{u-v} $$ if $u\neq v$. This is equivalent to $$ u\ln\frac uv > u-v\;\Longleftrightarrow\;\ln\frac uv > 1 - \frac vu. $$ Setting $x = v/u$, this is equivalent to $\ln x < x-1$ for $x\neq 1$, which we know is true.