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.