gcd and lcm – Carmichael number n such that $frac{gcd(n-1, phi(n))^2}{lambda(n)^2} geq n-1$.

Such that $phi(n)$ is a phi Euler function, $lambda(n)$ is a Carmichael lambda function.
With $frac{gcd(n-1, phi(n))^3}{lambda(n)^3} geq n-1$, i can find some numbers n such as: $1729, 19683001, 631071001, 4579461601, 8494657921$ and so on, but with $frac{gcd(n-1, phi(n))^2}{lambda(n)^2} geq n-1$, how can i find a number n??

Finite automaton for all words whose length $n$ satisfies $operatorname{gcd}(n,504) geq 6$

I have been working on the following homework question, and I just can’t seem to make any progress:

Construct a finite automaton having fewer than 36 states that recognizes the language ${s in a^* : operatorname{gcd}(|s|, 504) geq 6}$, where $|s|$ is the length of $s$.

So far I have been trying to figure out a regular pattern in $504$ and have broken it down to the prime factorization $504 = 2^3 cdot 3^2 cdot 7$, which means that all divisors must be multiples of 2, 3, or 7. However I don’t know how to create the finite automaton (NFA or DFA). Any help is appreciated!

inequality – Prove that $frac{b+c}{sqrt{a}}+frac{c+a}{sqrt{b}}+frac{a+b}{sqrt{c}} geq sqrt{a}+sqrt{b}+sqrt{c}$

If $a,b,c$ are positive reals and $abc=1$, Prove that
$$frac{b+c}{sqrt{a}}+frac{c+a}{sqrt{b}}+frac{a+b}{sqrt{c}} geq sqrt{a}+sqrt{b}+sqrt{c}$$
My try:
Is there a way from here?

Find a linear bounded automaton that accepts the language $L = { a^{n!} : n geq 0 }$

I need to construct linear bounded automaton for the language $L = { a^{n!} : n geq 0 }$. I know how LBA functions, however, I don’t have a thought how it can check the n! that to in the power of a. I might want to hear a few suggestions, as I am experiencing difficulty in developing the specific LBA for it.

$left(|x|^{p-2}x-|y|^{p-2}yright)(x-y) geq 0$

How can I prove this two statements:
For any $x,y in mathbb{R}$ we have

$left(|x|^{p-2}x-|y|^{p-2}yright)(x-y) geq 0$

and the second inequality

$left(|x|^{p-2}x-|y|^{p-2}yright) leq Cleft(|x|^{p-2}+|y|^{p-2}right)|x-y|$

Bound on $P(X-E(X) geq alpha)$

For a random variable $X$ with mean $mu$ and variance $sigma^2$ show that
$$P(X-nugeq alpha) leq frac{sigma^2}{sigma^2 + alpha^2} qquad alphageq 0$$

functional analysis – If $A$ is a self-adjoint operator then $(A(u),u) geq 0$?

Let $H=(H, (cdot, cdot ))$ be a Hilbert space over $mathbb{R}$ and $A : D(A) subset H longrightarrow H$, with $overline{D(A)}=H$, a linear self-adjoint operator. Suppose $ A $ has only a negative eigenvalue $lambda<0$, with eigenvector associated $v in D(A)$, then we have $A(v)=lambda v$.

Question. For all $w in D(A)$ such that $(w,v)=0$ we have
$$(A(u),u) geq 0?$$

What I got to think is: If $w in D(A)$ is such that $(w,v)=0$ then
$$0=(w,lambda v)=(w, A(v))= (A(w),v) $$
but I could not conclude anything.

linear algebra – Show there exists $pi in R_{++}^{nk}$ such that $A_{nk}(l)pi geq 0$ if and only if the determinant is non-negative

Let n,k be positive integers, with $n geq 2$ and $k geq 1$. Let l={${l_i}$} be an nk-dimensional positive real vector. Let $A_{n,k}(l)={(a_{i,j})}_{1 leq i,j leq nk}$ be the following $nk times nk$ matrix:
l_i quad if quad i=j \
-1 quad if quad 1 leq j-i leq n-1 \
-1 quad if quad i geq nk-n+2 quad and quad j+nk-i leq n-1 \
0 quad otherwise

I have following conjecture: $quad$
There exists $pi in R_{++}^{nk}$ such that $A_{nk}(l)pi geq 0$ if and only if the determinant of$A_{n,k}(l)$ is non-negative. But I do not know how to prove it. Any comments or suggestions would be greatly appreciated.

discrete mathematics – Prove that each tree $T=(V,E)$ with $|V| geq 2$ contains at least $2$ leaves.

I’m not sure if this lemma has a name or not but here goes:

Every tree $T=(V,E)$ with $|V| geq 2$ contains at least $2$ leaves.

The proof for this relies on the fact that if we start with only $2$ nodes, $v_1,v_2 in V$, then we have an $e in E$, which connects $v_1$ and $v_2$, then $T$ has $2$ leaves. If we have more than $2$ nodes, then we can pick either $v_1$ or $v_2$ and go in either direction. If we go in the direction of $v_1$ and it is not an end node, then we go to the succeeding node. If that is an end node, then we have one leaf. However, since we didn’t move in the direction of $v_2$, we have a total of $2$ leaves. We can repeat this for any number of nodes on either side.

However, how can we argue that it’s possible for there to be more than $2$ leaves?

real analysis – If $f(x) rightarrow infty$ as $x rightarrow a$, and $g(x) geq f(x)$ forall x near a. Does $g(x) rightarrow infty$ as $x rightarrow a$

I’m a student taking a real analysis course at university and I’m working down my problem sheet and I’ve been asked the question above.

I understand that if a function $f(x) leq g(x)$ then for all $x$ near $a$ by definition $g(x)$ has to be bigger at all times therefore it must tend to infinity if $f(x)$ does. I am under the impression we should use first principles. So

$$0 < |x-a|< delta implies f(x) > M $$

Is it enough just to say?

$$0 < |x-a|< delta implies f(x) leq g(x) > M $$

Thanks for your time! Any tips or tricks would be much appreciated!