A better lower bound for $f(t):=((t-a)^2+b)^2+d(t-c)^2$ on $[0,\infty[$ where $a,b,c,d>0$ and $c>a$ Let
$f(t):=((t-a)^2+b)^2+d(t-c)^2$ where $a,b,c,d>0$ and $c>a$, and  $t\geq 0$. I am trying to find a better lower bound for $f$ than $b^2$, if it exists.
The parabola $t \mapsto (t-a)^2+b$ has its unique minimum $b$ at $t=a$. The same applies to its square $((t-a)^2+b)^2$. Also, the parabola $t\mapsto d(t-c)^2$ has its unique minimum value $0$ at $t=c$.
If we ignore the term $(t-a)^4$, and look instead at $g(t):=2b(t-a)^2+d(t-c)^2+b^2$, the minimum can be shown to be $b^2$ which is obtained at
$t=(2ab+cd)/(2b+d)$. I don't know if this helps get a sharper answer.
Is there a way around computing the critical point of $f$ ?
 A: For a real-valued function, the maximum or minimum is taken where the first derivative is zero. To check whether there is a maximum or a minimum there is to test the sign of the second derivative.
For the set of four parameters as in the given function, the condition that there is an extremum is given by the following latex expression under the condition that $\{a,b,c,d\}$ are all reals.
$
\left\{\left\{t\to \text{ConditionalExpression}\left[\text{Root}\left[-2 a^3-2 a b^2-c d+\left(6 a^2+2 b^2+d \right) \text{ ss1}-6 a \text{ ss1}^2+2
\text{ ss1}^3\&,1\right],\left(9 a-9 c+\sqrt{6} \
\sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}>0\&\&-2 b+\sqrt{2} \
\sqrt{-d}>0\&\&2 b+\sqrt{2} \sqrt{-d}>0\&\&d<0\&\&-9
a+9 c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}>0\right)\
\left\|\left(9 a-9 c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}<0\&\&-2
b+\sqrt{2} \sqrt{-d}>0\&\&2 b+\sqrt{2} \sqrt{-d}>0\&\&d<0\right)\right \|\left(-2 b+\sqrt{2} \sqrt{-d}>0\&\&2 b+\sqrt{2} \
\sqrt{-d}>0\&\&d<0\&\&-9
a+9 c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}<0\right)\
\|\left(-2 b+\sqrt{2} \sqrt{-d}<0\&\&d<0\right)\|\left(2 b+\sqrt{2} \
\sqrt{-d}<0\&\&d<0\right)\|d>0\right]\right\},\left\{t\to
\text{ConditionalExpression}\left[\text{Root}\left[-2 a^3-2 a b^2-c \
d+\left(6 a^2+2 b^2+d \right) \text{ ss1}-6 a \text{ ss1}^2+2 \text{ ss1}^3\&,2\right],2
b+\sqrt{2} \sqrt{-d}>0\&\&-2 b+\sqrt{2} \sqrt{-d}>0\&\&9 a-9 \
c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}>0\&\&-9 a+9 \
c+\sqrt{6} \sqrt{-\frac{\left(2
b^2+d \right)^3}{d^2}}>0\&\&d<0\right]\right\},\left\{t\to \text{ConditionalExpression}\left[\text{Root}\left[-2 a^3-2 a b^2-c \
d+\left(6 a^2+2 b^2+d \right)
\text{ ss1}-6 a \text{ ss1}^2+2 \text{ ss1}^3\&,3\right],2 \
b+\sqrt{2} \sqrt{-d}>0\&\&-2 b+\sqrt{2} \sqrt{-d}>0\&\&9 a-9 \
c+\sqrt{6} \sqrt{-\frac{\left(2
b^2+d \right)^3}{d^2}}>0\&\&-9 a+9 c+\sqrt{6} \sqrt{-\frac{\left(2 \
b^2+d \right)^3}{d^2}}>0\&\&d<0\right]\right\}\right\}
$
The immense effort stems from the fourth degree of the function is.
Simply pluck the t-values and calculate the extremum.
$ss1$ is a free parameter for the calculation.
In case this is understood and welcome there be more of this kind of extra-long and large expressions for the general solution. To my part, I think is already fairly hard. And harder as the problem is formulated just for staying reals.
A: If we ignore the term $(t-a)^4$, and look instead at $g(t):=2b(t-a)^2+d(t-c)^2+b^2$, the minimum of $g$ can be shown to be $ bd+2bd(a-c)^2/(2b+d)+b^2$ which is obtained at
$t=(2ab+cd)/(2b+d)$.
