Let $A$ be $n\times n$ symmetric matrix, and $\lambda_1\geqq \cdots \geqq \lambda_n$ be eigenvalues of $A.$
And let $\{0\}\neq V$ be subspace of $\mathbb R^n$, $\dim V=k.$
Then, show that $\exists v\in V$ s.t. $\dfrac{(Av,v)}{(v,v)}\leqq \lambda_k$, where $(\cdot, \cdot)$ is Euclidean inner product.
My approach uses the property of symmetric matrix : diagonozable by orthogonal matrix．
There exists orthogonal matrix $U=(u_1, \cdots, u_n)$ s.t. $U^{-1}AU=\begin{pmatrix}\lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}, Au_i=\lambda_i u_i$ for $1\leqq i \leqq n.$
Then, I have $\dfrac{(Au_i, u_i)}{(u_i, u_i)}=\dfrac{\lambda_i (u_i, u_i)}{(u_i, u_i)}=\lambda_i$ for all $i=1,\cdots, n$.
I wonder how I should determine $v.$
If I determine $v=u_k$, I get $\dfrac{(Av,v)}{(v,v)} = \dfrac{(Au_k,u_k)}{(u_k,u_k)} =\lambda_k \leqq \lambda_k$ but I cannot check $v=u_k\in V$, and this doesn't use the magnitude correlation of $\lambda$'s ($\lambda_1\geqq \cdots \geqq \lambda_n$) so I think this $v$ doen't work.
Thanks for the help, which $v$ does work ?
As you have mentioned, if $u_k\in V$, then you are done. If $u_k\notin V$, then at least one among $u_{k+1},\ldots, u_n$ is in $V$. This is because $V$ has dimension $k$, so it cannot be contained in $\left<u_1, \ldots, u_{k-1}\right>$. Let $u_{k+h}\in V$, then $$\frac{(Au_{k+h}, u_{k+h})}{(u_{k+h}, u_{k+h})}=\lambda_{k+h}\leq \lambda_k.$$