Ytukyg

Let $mathcal{H}$ be a real Hilbert space and $a,b,c,d in mathcal{H}, t in (0,1)$ be such that
$$ leftlangle a - t cdot c , a - b rightrangle = (1-t) leftlangle b - t cdot d , a - b rightrangle + t leftlangle - t cdot d , a - b rightrangle neq 0 tag{1}label{1} . $$
Notice that if we set
$$ v := a-b, x:= b - t cdot d , y:= - t cdot d, z := a - t cdot c , $$
then we have
$$ leftlangle z , v rightrangle = (1-t) leftlangle x , v rightrangle + t leftlangle y , v rightrangle = leftlangle (1-t)x + ty , v rightrangle . $$
This reminds us about the convex combination
$$ z = (1-t)x + ty $$
in which we can get the following inequality
$$ leftlVert z rightrVert ^{2} leq (1-t) leftlVert x rightrVert ^{2} + t leftlVert y rightrVert ^{2} tag{2}label{2} . $$
The question is can we derive some similar inequality like eqref{2} for eqref{1}.
Otherwise, as a naive approach gives
begin{align*}
leftlVert a - t cdot c rightrVert ^{2} - leftlVert b - t cdot c rightrVert ^{2} & = (1-t) leftlVert b - t cdot d rightrVert ^{2} + t leftlVert - t cdot d rightrVert ^{2} - t (1-t) leftlVert b rightrVert ^{2} \
& qquad - leftlVert (2-t)b - t cdot d - a rightrVert ^{2} .
end{align*}
I would be happy to weaken the term $- leftlVert b - t cdot c rightrVert ^{2}$, that is to have $- delta leftlVert b - t cdot c rightrVert ^{2}$ for some $delta in (0,1)$.

Unfortunately, in this imperfect world we not always can be happy.

Simplifying each side in the last equality in your question we obtain

$$langle a,arangle-2tlangle a,crangle+t^2langle c,crangle-[langle b,brangle-2tlangle b,crangle+t^2langle c,crangle]le$$ $$langle a,arangle+2(2-t) langle a,brangle-2t langle a,drangle+(2t-3) langle b,brangle+2tlangle b,drangle.$$

As I understood you, it should be an inequality with some multiplier $delta$ before the square brackets, which is as small as possible. Then it is simplified to

$$0le (4-2t) langle a,brangle+2tlangle a,crangle-2t langle a,drangle+(2t-3+delta) langle b,brangle-2tdeltalangle b,crangle+2tlangle b,drangle+t^2(delta-1)langle c,crangletag{3}label{3}$$

On the other hand, (1) imply

$$ langle a,arangle+(t-2)langle a,brangle- tlangle a,crangle+t langle a,drangle+(1-t) langle b,brangle+tlangle b,crangle-tlangle b,drangle=0. tag{4}label{4}$$

Myltiplying (ref{4}) by two and adding to (ref{3}), we obtain

$$0le 2langle a,arangle+(delta-1) langle b,brangle+2t(1-delta)langle b,crangle+t^2(delta-1)langle c,crangle=$$ $$2|a,a|+(delta-1)|b-tc|.tag{5}label{5}$$

Now for any $delta<1$ we can easily find $a$, $b$, $c$, $d$, and $t$ satisfying (1), but violating (ref{5}). For the simplicity put $a=d=0$, $b=-cne 0$ and $t=1/2$.