Let $$ Ax=y, \quad ||y||=1 $$ where $A$ is an $n$ by $n$, symmetric PSD matrix, $0\leq \lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n $ are the eigenvalues.
1. Min-max theorem
Let $M=A^2$, so its $i$th eigenvalue is $ \lambda_i^2$. The min-max theorem
$$ \lambda_1 ^2 \leq \frac{x^TM x}{x^Tx} \leq \lambda_n^2 $$ and $$ x^TM x =\langle Ax, Ax \rangle = ||y||^2=1 $$ leads to $$ \frac{1}{\lambda_n^2 } \leq||x||^2\leq\frac{1}{\lambda_1^2 } \quad (1) $$
2. Cauchy-Schwarz inequality
Diagonalize $A=Q\Lambda Q^T$, so $A^{-1}=Q\Lambda^{-1} Q^T$, where $\Lambda^{-1}$ is a diagonal matrix containing $1/\lambda_i$.
Since $x=A^{-1}y$,
$$ ||x||^2=y^T (A^{-1})^T A^{-1} y= y^T Q (\Lambda^{-1})^2 Q^T y = |\langle \operatorname{diag}(\Lambda^{-1}),Q^Ty\rangle|^2 $$
let $z=Q^Ty$, we have $$ ||x||^2= \sum_i^n \frac{ z_i^2}{\lambda_i^2} $$
since $||Q^T y ||=||z||=1$, Cauchy-Schwarz inequality gives
$$ ||x||^2 \leq ||\operatorname{diag}(\Lambda^{-1}) ||^2 ||z||^2 =||\operatorname{diag}(\Lambda^{-1})||^2= \sum_i^n \frac{1}{\lambda_i^2} \quad (2) $$
3. problem
The result (1) is not consistent with (2), what's wrong? Especially, can $||x||^2$'s maximum value reach $$\frac{1}{\lambda_1^2}$$ or $$\sum_i^n \frac{1}{\lambda_i^2} $$?
First of all: in the sequence of equations, I believe that you meant to say $$ ||x||^2=y^T (A^{-1})^T A^{-1} y= y^T Q (\Lambda^{-1})^2 Q^T y = \langle\operatorname{diag}(\Lambda^{-1}),Q^Ty\rangle. $$ This statement is not quite true. Taking $z = Q^Ty$, we find that $$ z^T (\Lambda^{-1})^2 z = \frac 1{\lambda_1^2} z_1^2 + \cdots + \frac 1{\lambda_n^2}z_n^2. $$ From here, we could apply Hölder's inequality in the $p=1,q=\infty$ case to find that $\|x\|^2 \leq \frac{1}{\lambda_1^2}$.
Second: there is no inconsistency. As you established, we have $\frac{1}{\lambda_n^2 } \leq\|x\|^2\leq\frac{1}{\lambda_1^2 }$. This is consistent with the statement that $\|x\|^2 \leq \sum_{i=1}^n \frac{1}{\lambda_i^2}$.
So your statement derived from the misapplication of the Cauchy Schwarz inequality is correct.