If $f$ is continuous and non-negative, then $\inf (f)^2=\inf (f^2)$? I think that the following conjecture is true and I tried to proved as follows:

For any non-negative continuous function $f$ on $\left [ a,b \right ]$, let  $m(f)=\inf \left \{ f(t) : a\leq t\leq b  \right \}$. Then $m(f)^{2}=m(f^2)$.

My try to prove this:
Since $f$ is continuous on $\left [ a,b \right ]$, then $f$ has a minimum value, called $f(d)$ for some $d \in \left [ a,b \right ]$, therefore $m(f)^{2}=f(d)^{2}$.
Now, since $f$ is continuous, then $f^2$ is continuous and it admits a minimum value on $\left [ a,b \right ]$, called $f({d}')^{2}$, but then we have that $f(x)^{2}\geq f({d}')^{2}$ which implies $(f(x)+f(d'))(f(x)-f(d'))\geq 0$. But since $f$ is non-negative, $(f(x)+f(d'))\geq 0$, this implies that $f(x)-f(d')\geq 0$ for all $x$ on  $\left [ a,b \right ]$. This means that $f(d')$ is a minimum value of $f$, therefore $f(d')=f(d)$ and $m(f)^{2}=m(f^2)$.
Is this aproach correct?
 A: Your approach seems correct to me, but we should be able to prove the same will hold without the continuity restriction. Here's how I would frame the proof:
Let's suppose that there is such an $f$ such that $f(x) \geq 0$ on $[a,b]$ with $\inf_{[a, b]} f(x) = m$ but $\inf_{[a, b]} f(x)^2 \neq m^2.$ This implies that either there is some $t \in [a, b]$ such that $f(t)^2 \leq m^2,$ or that there is some tigher lower bound $n^2 > m^2$ such that $f(x)^2 \leq n^2$ for all $x \in [a,b].$
Regarding the first case, because the square root function is monotone decreasing we have that $f(t)^2 \leq m^2$ implies that $|f(t)| \leq |m|,$ and because both $f(t)$ and $m$ must be positive (if $m$ was negative then $0$ would be a tighter lower bound) this implies that $f(t) \leq m,$ which contradicts the definition of $m$ as a lower bound.
For the second case, similarly we have that $f(x)^2 \leq n^2$ implies that $f(x) \leq n$ for all $x \in [a,b]$ for some $n > m,$ which also contradicts the fact that $m$ is the infimum of $f(x)$ because it forms a tighter lower bound.
So, because $m^2$ is the highest possible lower bound of $f^2,$ it is the infimum of $f^2.$
(Note: your manipulation involving the difference of squares will also work in either case)
The non-negative condition cannot be dropped, as evidenced by the case of a function like $x^3$ on $[-1, 1],$ where $(\inf_{[-1, 1]} x^3)^2 = (-1)^2 = 1$ but $\inf_{[-1, 1]} (x^3)^2 = \inf_{[-1, 1]} x^6 = 0.$
