Strange SVD Bound with Frobenius Norm

Question

For any matrix $A$, show that $$ \sigma_k \le ||A||_F/\sqrt{k} $$ where $\sigma_k$ is the $k$-th singular value of $A$.

For $k=1$ I would say it's trivial, but for $k>1$?

Also tried this looking the link $$ \sigma_k=||Av_k||_2\le ||A||_F||v_k||_2 = ||A||_F $$ since $v_k$ is othonormal vector.

Already read other questions SVD Bound but can't find how to introduce that $k$ in the bound. Some hints?

Answer 1

It suffices to note that $$ \|A\|_F^2 = \sum_{i=1}^n \sigma_i^2 \geq \sum_{i=1}^k \sigma_i^2 \geq \sum_{i=1}^k \sigma_k^2 = k \sigma_k. $$