## inequalities – a simple functional inequality

Is there any general solution to the functional inequality f (x * y)? <= f(x) + f(y) . where x and y are in [0,1]? I can find many particular solutions but I just wonder if there is a general description of functions f satisfying such inequation. I have the following conditions 1) f : [0,1] -> R
2) f is not negative in [0,1]
2) f (1) is bounded
3) f is once continuously differentiable in (0,1)

Thanks for any suggestions

## Norma – Inequality between two sums.

As part of a research problem that I am working on, I need to show the following inequality. Leave $$x = (x_1, dots, x_K)$$ with $$x_i> 0$$ for all $$i$$. So, I want to show that
$$frac {1} {K ^ 2} sum_ {i = 1} ^ K x_i geq K left ( sum_ {i = 1} ^ K x_i ^ {- 1/2} right) ^ { -two}.$$
by $$K = 2$$ This is easy to show with simple algebra, but in the general case I could not find a proof. Note that this inequality can be rewritten as
$$sum_ {i, j, k} ^ K frac {x_k} { sqrt {x_ix_j}} geq K ^ 3.$$
Numerically it always seems to hold clearly. Using the fact that the $$1/2$$-norma is higher than the $$1$$-norma, could show that it is higher than $$K ^ 2$$, but none $$K ^ 3$$.

Any help would be greatly appreciated!

## Inequality simpmodulus

I want to determine everything $$z in mathbb C$$ for which
$$begin {equation} tag {1} label {1} left | frac {2 + z} {2-z} right | le 1. end {equation}$$
In fact, I already know that eqref {1} is equivalent to $$Re (z) le0$$ (where $$Re (z)$$ denotes the real part of $$z$$.) However, I do not know how to get to this result and I do not know how to get results for more general inequalities, such as $$dots le c$$ by some constant instead of $$1$$.

## Show the inequality \$ sec x ge 1+ (x ^ 2/2) \$

Test inequality

$$sec (x) ge 1+ (x ^ 2/2)$$

## resolving a hard inequality

by $$a, b, c> 0$$I have to prove that $$sum_ {cyc} ^ {} frac {a (a ^ 3 + b ^ 3)} {a ^ 2 + ab + b ^ 2} ge frac {2} {3} (a ^ 2 + b ^ 2 + c ^ 2).$$

We have:

$$sum_ {cyc} ^ {} frac {a (a ^ 3 + b ^ 3)} {a ^ 2 + ab + b ^ 2} = sum_ {cyc} ^ {} frac {a ^ 4 } {a ^ 2 + ab + b ^ 2} + sum_ {cyc} ^ {} frac {ab ^ 3} {a ^ 2 + ab + b ^ 2} ge frac {(a ^ 2 + b ^ 2 + c ^ 2) ^ 2} {a ^ 2 + b ^ 2 + c ^ 2 + ab + bc + ca} + sum_ {cyc} ^ {} frac {ab ^ 3} {a ^ 2 + ab + b ^ 2} ge frac {1} {2} (a ^ 2 + b ^ 2 + c ^ 2) + sum_ {cyc} ^ {} frac {ab ^ 3} {a ^ 2 + ab + b ^ 2}$$

but I can not keep going

## Theory of complexity: Contradiction test for the inequality of P and NP?

I am trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after the deduction, he said that this is problematic where I can not find a convincing reason to accept.

We begin by assuming that $$P = NP$$. Then yield that $$SAT in P$$ which in turn follows that $$SAT in TIME (n ^ k)$$. As stands, we are able to reduce all the languages ​​in $$NP$$ to $$SAT$$. Thus, $$NP subseteq TIME (n ^ k)$$. On the contrary, the time hierarchy theorem states that there should be a language $$A in TIME (n ^ {k + 1})$$, that's not in $$TIME (n ^ k)$$. This would lead us to conclude that $$A$$ is in $$P$$, while not in $$NP$$, which is a contradiction to our first assumption. So, we came to the conclusion that $$P neq NP$$.

Is there something wrong with my test? I was struggling for hours before asking this, however!

## Algorithms – Find complexity by solving inequality

I would like to calculate the complexity of an algorithm using the "inequality strategy".

This algorithm takes two integers as one entry. $$n, m$$. In addition this algorithm is recursive and if we denote $$C_ {n, m}$$ The number of operations that this algorithm performs for the entry. $$n, m$$ We have the following inequality:

$$C_ {n, m} leq C_ {n, (m-1)} + C _ (n-1), (m-1)} + O (1)$$

In addition, the algorithm ends for the entry: $$(0, k)$$ Y $$(k, 0)$$ for all $$k in mathbb {N}$$ Y $$C_ {0, k} = O (1)$$ Y $$C_ {k, 0} = O (1)$$.

So now I need to somehow solve the inequality to get $$C_ {n, m}$$ for all $$n, m$$. However, I really do not know how to do this.

The problem is that I think that any function works. For example, if I say: $$C_ {n, m} = O (m)$$ then it is tre since by unduction we have then:

$$C_ {n, m} leq O (m-1) + O (m-1) + O (1) = O (m-1)$$

Also if I say that: $$C_ {n, m} = O (n)$$ Then it also works since by induction:

$$C_ {n, m} leq O (n) + O (n-1) + O (1)$$

So, what is the problem with what I am doing and how to solve this inequality to obtain the complexity of $$C_ {n, m}$$ ?

Thank you !

## pr.probabilidad – inequality of Hanson-Wright – MathOverflow

As indicated in this document, a variant of the Hanson-Wright inequality is the following:

Leave $$A$$ be a fixed $$n times n$$ matrix. Consider a random vector $$X = (X_1, ldots, X_n)$$ where $$X_i$$ They are independent random variables that satisfy $$E[X_i] = 0$$Y $$|| X_i || _ { psi_ {2}} leq K$$. So for any $$t geq 0$$, we have
begin {align} P big (X ^ {T} AX – E[X^{T}AX] > t big) leq 2 exp[-c min{dfrac{t^2}{K^4 ||A||_F^2}, dfrac{t}{K^2 ||A||}}]. end {align}

In this equation, $$|| A || _F$$ It's the Frobenius standard. $$A$$ Y $$|| A ||$$ is the spectral norm of $$A$$.

Are there results where instead of $$|| A || _F ^ 2$$ Y $$|| A ||$$, we have $$|| A || _F$$ Y $$|| A || ^ {1/2}$$?

## shows this inequality with \$ xyzw = 1 \$.

leave $$x, y, z, w> 0$$,such $$xyzw = 1$$,show that
$$3 (x + y) (y + z) (z + w) (w + x) ge 16 (x + y + z + w-1)$$

I have three variable inequalities

## Inequality – What is the value of k?

Find all the real value of $$k$$ such that the equation system
begin {align} a ^ 2 + ab & = kc ^ 2 \ b ^ 2 + bc & = ka ^ 2 \ c ^ 2 + ca & = kb ^ 2 end {align}
have a positive solution of real numbers for $$a$$, $$b$$Y $$c$$.

I have already found the value of $$k$$. Here is my solution.
Adding the whole equation, we find that
$$a ^ 2 + b ^ 2 + c ^ 2 + ab + bc + ca = k left (a ^ 2 + b ^ 2 + c ^ 2 right)$$
or
$$k – 1 = frac {ab + bc + ca} {a ^ 2 + b ^ 2 + c ^ 2}.$$
As $$ab + bc + ca leq a ^ 2 + b ^ 2 + c ^ 2$$ for $$a, b, c in mathbb {R ^ {+}}$$,
$$k – 1 leq frac {ab + bc + ca} {ab + bc + ca} = 1 iff k leq 2.$$
My question is, is my solution incorrect? Or did I miss something? Thank you.