Convex inequality with the function $f(x)=x\log x$

Question

Let $u(t):=t\log t$. This function is convex on $[0,+\infty)$ and $u(0)=0$. How can I prove that$$u(t+h)-u(t)\geq u(h)\quad\forall t,h\in[0,+\infty)?$$ Since $u(0)=0$ using convexity we know that $u(st)\leq s\,u(t)$, where $s\in[0,1]$ but i get confused with the inequalities.

Answer 1

Note, that

$$\begin{array}{rcl} u(t)+u(h)&=&u((t+h)\dfrac{t}{(t+h)})+u((t+h)\dfrac{h}{(t+h)}) \\ &\leq& \dfrac{t}{(t+h)}\cdot u(t+h)+\dfrac{h}{(t+h)}\cdot u(t+h)=u(t+h) \quad \forall t,h\in[0,+\infty) \end{array}$$

Answer 2

It is one line (\log is monotone):

$$u(t+h)=(t+h)\log(t+h)=t\log(t+h)+h\log(t+h)\ge t\log(t)+h\log(h)=u(t)+u(h).$$