Let $h:[0,1]\to \mathbb{R}$ be continuous. Prove that there exists $w\in [0,1]$ such that

$h(w)=\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left(\frac{1}{2}\right)+\frac{w+1}{12}h(1).$

I tried several things with this problem. I first tried a new function

$\begin{array}{rl}g(x)& =h(x)-\frac{x+1}{2}h(0)+\frac{2x+2}{9}h\left(\frac{1}{2}\right)+\frac{x+1}{12}h(1)\\ & =h(x)-(x+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)).\end{array}$

Evaluating $g(0)$ and $g(1)$ didn't really work.

Since I want to find a point ${x}_{0}$ where $g({x}_{0})>0$ and another ${x}_{1}$ where $g({x}_{1})<0$, I started thinking how I can make $g({x}_{0})>0$,or

$h({x}_{0})-({x}_{0}+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1))>0,$

or

$\frac{h({x}_{0})}{{x}_{0}+1}>\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1).$

Since the LHS is the gradient from the point $(-1,0)$ to $({x}_{0},h({x}_{0}))$, I need to somehow prove that that

$\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)$

is always less than the steepest gradient from $(-1,0)$ to $({x}_{0},h({x}_{0}))$. However, I am stuck here. Any help?