Showing $f(x) \geq 2f(\frac{u+v}{2})-f(\frac{u+v}{2}-t)$ for $f$ convex

Question

consider a convex function $f$ on a compact subinterval $[u,v]$.

write an arbitrary point $x\in [u,v]$ as $x=(u+v)/2+t$ for some $t$ with $\vert t\vert \leq (v-u)/2$.

Then $f(x)=f(\frac{u+v}{2}+t)$, but how to go from here to

$$ f(x)=f(\frac{u+v}{2}+t)\geq 2f(\frac{u+v}{2})-f(\frac{u+v}{2}-t) $$

Im sure it follows from writing $\frac{u+v}{2}+t$ as $2\frac{u+v}{2} - (\frac{u+v}{2}-t)$ and then somehow doing a convex combination, and using the definition of convexity. I don't see what convex combination works though.

Answer 1

$$ \frac{u+v}{2} = \frac 12 \left( \left(\frac{u+v}{2} - t \right) + \left(\frac{u+v}{2} + t \right) \right) $$ implies for the convex function $f$ that $$ f\left(\frac{u+v}{2}\right) \le \frac 12 \left( f\left(\frac{u+v}{2} - t \right) + f\left(\frac{u+v}{2} + t \right) \right) $$