Suppose I have cubic functions $$f(t)=a_0+a_1t+a_2t^2+a_3t^3$$ and $$g(t)=b_0+b_1t+b_2t^2+b_3t^3$$ where $t\geq 0$.
Let $$h(t)=f(\alpha t)+g((1-\alpha)t), \alpha\in[0,1].$$ What conditions do I impose on the coefficients of $f$ and $g$ for $$(1)\space{} h(t)\geq f(t)?$$ $$(2)\space{} h(t)\geq g(t)?$$
Initially, in (1) I considered that $h(t)-f(t)\geq 0$. But I got stuck because I cannot infer from there that the coefficients of $h(t)-f(t)$ would each be greater than or equal to zero, right? Is there a special theory that needs to be used to solve this kind of problem?
HINT.-Notice that $\alpha t+(1-\alpha)t=t$ so the restriction for $\alpha\in [0,1]$ is not a restriction for $t\in\Bbb R$. You want to have $$\begin{cases}h(t)-f(t)\ge 0\\h(t)-g(t)\ge 0\end{cases}$$
Excepting the case $a_3=b_3$ you have these two equations are cubics so they can not be always non negative. Hence it is necessary both equation be of the second degree
It follows $\color{red}{a_3=b_3}$ is a first condition. The second one is easy to get for quadratic equation (the minimum must be greater or equal than zero which involve a straightforward calculation).