Proving a certain equality only holds for $p=2$ While solving a problem I arrived at a specific condition which I am pretty sure is true but I am having a hard time proving it. Let $0<t\leq 1$ and $p\geq 1$ be real numbers satisfying the following equality
$$
2^{2/p}+(2-2^p+2^p t^p)^{2/p}=4t^2
$$
Additionally, $2-2^p+2^p t^p\geq 0$. I want to show that this can only happen when $p=2$. I suspect this has to do with some inequalities between $l_p$ norms, but I can't quite figure it out.
 A: By the generalized mean inequality we have that if $r<s$ then $$\left(\frac{a^r+b^r}2\right)^{1/r} \leq \left(\frac{a^s+b^s}2\right)^{1/s}$$ for all $a,b\geq 0$.
The equality $4t^{2}=2^{2/p}+\left(2-2^{p}+2^{p}t^{p}\right)^{2/p}$ can be written as $$\left[2t^{2}\right]^{p/2}=\left[\frac{2^{2/p}+\left(2-2^{p}+2^{p}t^{p}\right)^{2/p}}{2}\right]^{p/2}$$ that reduces to $2t^2=2t^2$ when $p=2$.
Suppose that $p<2$. Then $2/p>1$ and by the generalized mean inequality we have
$$
\begin{align}
\left[2t^{2}\right]^{p/2} &= \left[\frac{2^{2/p}+\left(2-2^{p}+2^{p}t^{p}\right)^{2/p}}{2}\right]^{p/2} \\
&\geq \frac{2+\left(2-2^{p}+2^{p}t^{p}\right)}{2} \\
&= 2-2^{p-1}\left(1-t^{p}\right) \\
&\geq 2t^{p} \\
\end{align}
$$
where the last inequality holds since $p<2$ and $1-t^p>0$ (and since $2-2^{x-1}(1-t^p)$ is a decreasing function in $x$ in this case).
Since $\left[2t^{2}\right]^{p/2} \geq 2t^p$ implies $2t^{2}\geq2^{2/p}t^{2}$ and $t>0$, we get $2\geq2^{2/p}$, a contradiction since $2/p>1$.
Suppose that $p>2$. All previous inequalities are reversed getting another contradiction at the end.
Therefore the only possible value for $p$ is $2$.
