Let $ f: [0,1] \to \mathbb R $ be a $ C^2 $ function such that $ f^{''} (t) \geq K $, where $ K$ is a real constant.
Can we write the above differential inequality equivalent in an integral inequality ?
$$ f(t) \leq (1- \lambda) f(0) + \lambda f(1) + \frac{K t(1-t)}{2} $$
for all $ \lambda, t \in [0,1]$.
After the replies it seems that we can't obtain such an integral bound, so my new question is if it possible to have such an equivalent way of writting the differential inequality $ f^{''} \geq K$ if we change sign to the last term, i.e.
New question: Is it true that
$$ f^{''}(t) \geq K \Longleftrightarrow f(t) \leq (1- \lambda) f(0) + \lambda f(1) - \frac{K t(1-t)}{2} \quad \forall \lambda, t \in [0,1] $$
Thank you in advance.
edit: Add a new question.
If $f$ and $g$ are two integrable functions on $[a, b]$ with $f(x) \le g(x)$ for all $x \in [a, b]$, then $$\int_a^b f(x)\ dx \le \int_a^b g(x)\ dx.$$ In particular, if $F' = f$ and $G' = g$ and $F(a) \le G(a)$, then $F(x) = \int_a^x f(t)\ dt + F(a)$ and $G(x) = \int_a^x g(t)\ dt + G(a)$. So $F(x) \le G(x)$ everywhere in $[a, b]$. From which it follows that $$\int_a^b F(x)\ dx \le \int_a^b G(x)\ dx$$ as well.
(Edited to correct that I originally had the inequality reversed.)
To apply this to your problem: $f'' \ge K$ implies $$f'(x) - f'(0) = \int_0^x f''(t)\ dt \ge \int_0^x K\ dt = Kx$$ $$f'(x) \ge Kx + f'(0)$$ And therefore, $$f(t) - f(0) = \int_0^t f'(x)\ dx \ge \int_0^t (Kx + f'(0))dx = \frac{Kt^2}{2} + f'(0)t$$ $$f(t) \ge \frac{Kt^2}{2} + f'(0)t + f(0)$$ for all $t \in [0,1]$ (or other interval in which $f'' \ge K$).
As user58697 has pointed out, in your inequality which you hoped is true for all $t, \lambda \in [0,1]$, if we set $t = 0, \lambda = 1$, we get $f(0) \le f(1)$. This is certainly not true of every function of bounded 2nd derivative on [0,1]. If you want a specific counterexample, let $f(x) = 1 - x$.