Intermediate Value theorem inequality problem Let $h:[0,1]\rightarrow \mathbb{R}$ 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{align*}
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)\left(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)\right).
\end{align*}
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 $$h(x_0)-(x_0+1)\left(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)\right)>0,$$
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?

Let $g:[0,1]\rightarrow \mathbb{R}$ be defined by
$$
g(w)=h(w) - \left(\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{w+1}{12}h(1)\right)
$$
It suffices to show that $g(0),g(1/2),g(1)$ cannot all be positive, and cannot all be negative.$\\[9pt]$
\begin{align*}
\text{Let}\;\, a &= h(0),\;b=h(1/2),\;c=h(1)\\[9pt]
\text{Then}\;\,g(0) &= h(0) - \left(\frac{1}{2}h(0)+\frac{2}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{1}{12}h(1)\right)\\[0pt]
&=a - \left(\frac{1}{2}a+\frac{2}{9}b+\frac{1}{12}c\right)\\[0pt]
&=\frac{1}{2}a-\frac{2}{9}b-\frac{1}{12}c\\[9pt]
g\left({\small{\frac{1}{2}}}\right) &= h\left({\small{\frac{1}{2}}}\right) - \left(\frac{3}{4}h(0)+\frac{1}{3}h\left({\small{\frac{1}{2}}}\right)+\frac{1}{8}h(1)\right)\\[0pt]
&=b - \left(\frac{3}{4}a+\frac{1}{3}b+\frac{1}{8}c\right)\\[0pt]
&=-\frac{3}{4}a+\frac{2}{3}b-\frac{1}{8}c\\[9pt]
g(1) &= h(1) - \left(h(0)+\frac{4}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{1}{6}h(1)\right)\\[0pt]
&=c - \left(\frac{1}{2}a+\frac{2}{9}b+\frac{1}{12}c\right)\\[0pt]
&=-a-\frac{4}{9}b+\frac{5}{6}c\\[9pt]
\text{Let}\;\,{\mathbf u} &= \langle g(0),g(1/2),g(1) \rangle\\[4pt]
{\mathbf v} &= \langle 18,8,3 \rangle\\[9pt]
\text{Then}\;\,
{\mathbf u}
\cdot
{\mathbf v}
&=
\langle g(0),g(1/2),g(1) \rangle
\cdot
\langle 18,8,3 \rangle\\[6pt]
&=
(18)g(0)
+
(8)g(1/2)
+
(3)g(1)
\\[4pt]
&=
(18)\left(\frac{1}{2}a-\frac{2}{9}b-\frac{1}{12}c\right)
+
(8)\left(-\frac{3}{4}a+\frac{2}{3}b-\frac{1}{8}c\right)
+
(3)\left(-a-\frac{4}{9}b+\frac{5}{6}c\right)
\\[4pt]
&=0\qquad\text{[identically]}
\end{align*}
It follows that $g(0),g(1/2),g(1)$ cannot all be positive, and cannot all be negative, as was to be shown.
