inequality – Probability that a system of linear inequalities of random variables is satisfied

Let $$X_i$$ (i = 1…n) be a set of independent uniform random variables. Is there a tool/method/concept to find the probability that a system of linear inequalities of $$X_i$$ is satisfied ?

$$left{ begin{array}{c} a_1 < b_{11} X_1 + b_{12} X_2 + … + b_{1n} X_n < c_1 \ … \ a_j < b_{j1} X_1 + b_{j2} X_2 + … + b_{jn} X_n < c_j \ … \ a_m < b_{m1} X_1 + b_{m2} X_2 + … + b_{mn} X_n < c_m end{array} right.$$
with $$a_j, b_{ji}, c_j in R$$

For a single inequality, one can find the probability using convolution.

For several inequalities and $$n < 4$$, one can find the probability using a geometric method.

But what if $$n geqslant 4$$ ?

inequality – IF \$a,b,c,d,e\$ : Real number other than \$-2≤a≤b≤c≤d≤e≤2\$,Prove the following equation \$frac{1}{b-a}+frac{1}{c-b}+frac{1}{d-c}+frac{1}{e-d}≥4\$

Question
IF $$a,b,c,d,e$$ : Real number other than $$-2≤a≤b≤c≤d≤e≤2$$,Prove the following equation
$$frac{1}{b-a}+frac{1}{c-b}+frac{1}{d-c}+frac{1}{e-d}≥4$$

I tried to use Cauchy-Schwarz like
$$(frac{1}{b-a}+frac{1}{c-b}+frac{1}{d-c}+frac{1}{e-d})(b-a+c-b+d-c+e-d)≥16$$
I think I should prove it using -2≤a≤b≤c≤d≤e≤2, but I’m confused about how to do it. Can anyone help me?

pr.probability – An Inequality of Expected Value of Random Variables

I encountered the following problem in my research:

Suppose there are $$N$$ random variables that are independent and identically distributed (IID). The probability density function (PDF) of these random variables $$f(x)$$ is a unimodal function symmetrical about $$0$$ (i.e., $$f(x)$$ is non-decreasing within $$(-∞,0)$$, and for any $$x$$, $$f(x) = f(-x)$$ holds. for example, the distribution can be uniform distribution, normal distribution, Cauchy distribution with mean $$0$$, etc.).
For a given real number $$x_0$$, Sort these random variables as $$X_1, X_2, …, X_N$$ such that $$|X_1-x_0|leq |X_2-x_0| leq … leq |X_N-x_0|$$
For example, if $$N = 3$$, the $$N$$ random variables are randomly chosen as $$-0.5, 1.5, 5$$, and $$x_0 = 1$$, then $$X_1 = 1.5, X_2 = -0.5, X_3 = 5$$.
Let $$Y_i = |frac{X_1+X_2+…+X_i}{i}-x_0|^r (i=1,…,N, r = 1 or 2)$$, then for any $$x_0$$ and $$f(x)$$, does the inequality
$$EY_1 leq EY_2 leq… leq EY_N$$
always hold? Where $$E$$ denotes the expected value.

The inequality above is tested via the Monte Carlo method for cases where the distributions are uniform distribution, normal distribution, and Cauchy distribution. Details can be seen in https://math.stackexchange.com/questions/4039555/an-inequality-of-expected-value-of-random-variables since I cannot post figures here…

Moreover, is it possible to derive the PDF of $$Y_i$$?

Answers or ideas for either $$r=1$$ or $$r=2$$ would be so grateful!

inequality – When can we deduce \$(mA+nB+pC)>(mX+nY+pZ)\$ from \$(A+B+C)>(X+Y+Z)\$?

I am trying to prove an equality in the form $$(mA+nB+pC)>(mX+nY+pZ)$$.
I can prove the inequality $$(A+B+C)>(X+Y+Z)$$ and I wonder if there is any condition,
under which we can deduce $$(mA+nB+pC)>(mX+nY+pZ)$$ from $$(A+B+C)>(X+Y+Z)$$ ?

And $$A,B,C,X,Y,Z,m,n,p$$ are all positive.

The first thing comes to mind is the Cauchy-Schwarz Inequality:

$$sqrt{left(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)}geleft(mX+nY+pZright)$$

Now it suffices to prove that

$$left(mA+nB+pCright)gesqrt{left(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)}$$
$$iffleft(mA+nB+pCright)^{2}geleft(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)$$

But I can’t find the condition under which we can deduce the above inequality from $$(A+B+C)>(X+Y+Z)$$.

Or using Hölder’s inequality:

$$left(mA+nB+pCright)^{frac{1}{2}}left(frac{1}{m}+frac{1}{n}+frac{1}{p}right)^{frac{1}{2}}ge A^{frac{1}{2}}+B^{frac{1}{2}}+C^{frac{1}{2}}$$
$$iffleft(mA+nB+pCright)^{frac{1}{2}}geleft(A^{frac{1}{2}}+B^{frac{1}{2}}+C^{frac{1}{2}}right)left(frac{1}{m}+frac{1}{n}+frac{1}{p}right)^{-frac{1}{2}}$$

And so on.

Proving inequality on \$mathbb{R}^+\$

Suppose, WLOG, $$a < b.$$

Prove that ineuqality $$a^b + b^a < a^a + b^b$$ is true for any $$a, b in mathbb{R}^+.$$

I need an elementary proof, without or with minimal use of elementary calculus. Thanks!

co.combinatorics – Matrix inequality \$a X succeq arcsin(X)\$ for some \$a > 0\$

Let $$X in S^{n}_{+}$$ be a positive semi-definite matrix with $$X_{ii} = 1$$ for all $$i leq n$$ (thus $$X$$ is a correlation matrix).
Since $$X$$ is positive semi-definite, we have $$|X_{ij}| leq 1$$ for any entry $$(i,j)$$.

Let $$arcsin(X) = (arcsin(X_{ij}))_{i,j leq n}$$, i.e., the arcsin function applies on the matrix X element-wisely. From the Schur product theorem, we know $$arcsin(X) succeq X$$, where $$A succeq B$$ means $$A-B$$ is positive semi-definite. I wonder, do we have an inequality on the other side, i.e., there exists some real number $$a> 0$$ such that $$a X succeq arcsin(X)$$?

I am asking this question because SDP relaxation usually provides good approximation algorithms to combinatorial optimization with a maximization objective. I am thinking about if we can still apply SDP to minimization problems.

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:
$$frac{b+c}{sqrt{a}}+frac{c+a}{sqrt{b}}+frac{a+b}{sqrt{c}}=frac{b+c+a-a}{sqrt{a}}+frac{c+a+b-b}{sqrt{b}}+frac{a+b+c-c}{sqrt{c}}$$
$$implies$$
$$frac{b+c}{sqrt{a}}+frac{c+a}{sqrt{b}}+frac{a+b}{sqrt{c}}=(a+b+c)(frac{1}{sqrt{a}}+frac{1}{sqrt{b}}+frac{1}{sqrt{c}})-(sqrt{a}+sqrt{b}+sqrt{c})$$
Is there a way from here?

if \$a×b×c<180\$ , then how many possible positive integer solution are there for the inequality?

As most of you know there is a classical question in elementary combinatorics such that if $$a×b×c$$=180 , then how many possible positive integer solution are there for the equation ?

The solution is easy such that $$180=2^2×3^2×5^1$$ and so , for $$a=2^{x_1}×3^{y_1}×5^{z_1}$$ , $$b=2^{x_2}×3^{y_2}×5^{z_2}$$ , $$c=2^{x_3}×3^{y_3}×5^{z_3}$$ .

Then: $$x_1+x_2+x_3=2$$ where $$x_i≥0$$ , and $$y_1+y_2+y_3=2$$ where $$y_i≥0$$ and $$z_1+z_2+z_3=1$$ where $$z_i≥0$$.

So , $$C(4,2)×C(4,2)×C(3,1)=108$$.

Everything is clear up to now.However , i thought that how can i find that possible positive integer solutions when the equation is $$a×b×c<180$$ instead of $$a×b×c=180$$

After , i started to think about it. Firstly , i thought that if i can calculute the possible solutions for $$x_1+x_2+x_3<2$$ where $$x_i≥0$$ , and $$y_1+y_2+y_3<2$$ where $$y_i≥0$$ and $$z_1+z_2+z_3<1$$ where $$z_i≥0$$ , then i can find the solution.However , there is a problem such that when i calculate the solution , i do not include the prime numbers and their multiplicites which is in $$180$$.

For example , my solution does not contain $$1×1×179<180$$

My question is that how can we solve these types of question . Is there any TRICK for include all possible ways ? Moreover ,this question can be generalized for $$a×b×c≤180$$ , then what would happen for it ?

inequality – Is \${nchoose k}2^ngeq{2nchoose k}\$?

Given $$n$$ and $$k$$ where $$kleq n$$. Is $${nchoose k}2^ngeq{2nchoose k}$$?

My progress so far: We know that $$big(frac{n}{k}big)^kleq{nchoose k}leqbig(frac{en}{k}big)^k$$. Applying this fact, we have $${nchoose k}2^ngeqbig(frac{n}{k}big)^k2^n$$ and $${2nchoose k}leqbig(frac{2en}{k}big)^k$$. So $${nchoose k}2^ngeq{2nchoose k}$$ holds when $$big(frac{n}{k}big)^k2^ngeqbig(frac{2en}{k}big)^kiff kleqfrac{nlog 2}{log(2e)}$$. This condition is quite close to what I desire ($$kleq n$$), but not exactly it.

systems of equations – Set of solutions for given inequality

Given the matrix $$Ainmathbb{R}^{ntimes n}$$ with all eigenvalues inside the unit circle and the symmetric positive definite matrix $$Pinmathbb{R}^{ntimes n}$$ satisfying $$APA^top-P+I=0$$, I need to find the set of solutions $$S$$ for

$$S P^{-1} S^top + AS^top + S A^top leq 0$$

It is obvious that this set of solutions is nonempty since $$S=0$$ satisfies the above relation. Also another trivial solution is $$S=-A P$$ as well as $$S= -alpha v v^top$$ for some values of $$alpha>0$$, where $$v$$ is an eigenvector of $$A$$. However, I wanted to see if there is any more general form of the solution $$S$$.