Matrix norm of $A-B$ and their smallest eigenvalues Let $A$ and $B$ be symmetric positive definite matrices of size $n\times n$
such that $\|A - B\| < \frac{1}{4}\lambda_{\min}(B)$
where $\|\cdot\|$ is the matrix 2-norm
and $0 < \lambda_{\min}(B)$.
[An additional condition is added]
For any $|A_{ij} - B_{ij}| < \frac{\lambda_{\min}(B)}{4n}$ for all $1\le i, j \le n$.
Can we say that 
$$
(*)\quad \lambda_{\min}(A) \ge \frac{3}{4}\lambda_{\min}(B)?
$$
If them, how so?
Since $\|B\| = \lambda_{\max}(B)$, it seems that the following is true
$$
\frac{3}{4}\lambda_{\min}(B) \le \lambda_{\max}(B)-\frac{1}{4}\lambda_{\min}(B)=\|B\|-\frac{1}{4}\lambda_{\min}(B) < \|A\| 
$$
however, not sure why the above (*) is true...and how to relate the smallest eigenvalue of $A$ out of this.
Any suggestions/comments/answers will be very appreciated.
 A: Yes, we can say it. By spectral theorem, we can write $A=PDP^T$ where $P$ is orthogonal and $D>0$ is diagonal with each entry being eigenvalue of $A$. We observe that 
$$
\min_{\|x\|=1}x^TAx=\min_{\|x\|=1}(P^Tx)^TD(P^Tx)=\min_{\|y\|=1} y^TDy=\lambda_{\text{min}}(A).\tag{*}
$$ ($(*)$ is a characterization of $\lambda_{\text{min}}(A)$.) So it suffices to show that $
\min_{\|x\|=1}x^TAx\ge \frac34 \lambda_{\text{min}}(B),
$ or equivalently $x^TAx\ge \frac34 \lambda_{\text{min}}(B)$ for all unit vectors $x$. We find that
$$
x^TAx=x^T(A-B)x+x^TBx.
$$ For the first term, we have by Cauchy-Schwarz, 
$$
-x^T(A-B)x\le|\langle x,(A-B)x\rangle|\le \|x\|\|(A-B)x\|\le \|A-B\|\le \frac14\lambda_{\text{min}}(B)
$$ and for the second term, by $(*)$,
$$
x^TBx\ge \lambda_{\text{min}}(B).
$$ Combining them, we get
$$
x^TAx=x^T(A-B)x+x^TBx\ge \frac34\lambda_{\text{min}}(B)
$$ as wanted. So $\lambda_{\text{min}}(A)\ge \frac34\lambda_{\text{min}}(B)$.
Note: The argument essentially shows that in general,
$$
\lambda_{\text{min}}(A)\ge \lambda_{\text{min}}(B)-\|A-B\|.
$$ In fact, in the same manner, we can also prove
$$
\lambda_{\text{max}}(A)\le \lambda_{\text{max}}(B)+\|A-B\|
$$ by using similar characterization $\max_{\|x\|=1}x^TAx=\lambda_{\text{max}}(A).$
