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### inequality (a quickie)

Posted: January 10th, 2010, 1:29 am
also in a form harmonic-like.prove:for any real sequence x_i (i=1, 2, ..., n)

### inequality (a quickie)

Posted: January 10th, 2010, 7:19 pm

### inequality (a quickie)

Posted: January 11th, 2010, 1:53 pm
Very nice! Where is it taken from?

### inequality (a quickie)

Posted: January 11th, 2010, 2:38 pm

### inequality (a quickie)

Posted: January 11th, 2010, 3:12 pm
Can you explain the first equality please? EDIT: is it seriously your contention that???In one you are carrying the sum over the integrals. In the other you have part of the integrand outside the sum; in other words, you still have j and k dependence in the final valueYour method is salvageable, at least for cases where you are allowed to exchange infinite sums and integrals, but it's not obvious that this is always the case.

### inequality (a quickie)

Posted: January 11th, 2010, 3:23 pm
zerdna, you cannot take the factor out of the double sum What we have is rather

### inequality (a quickie)

Posted: January 11th, 2010, 3:27 pm
sorry, i am very slow with latex. DJ corrected me before i corrected myself. The idea is valid though.

### inequality (a quickie)

Posted: January 11th, 2010, 3:32 pm
nice solutions! it's a 30-sec challenge from a friend. his "official" solution kinda coincides with DJAverage's:by changing variable from "t" to exp(-t), one should get zerdna's proof (with the above correction).another way to interpret this is to prove matrix C_{ij}=1/(i+j) is a covariance matrix. can you find a set of random variables resulting this covariance matrix?

### Re: inequality (a quickie)

Posted: September 20th, 2019, 3:58 pm
Let [$]x,y,z[$] be positive numbers. Prove that:

[$]\sqrt[3]{\frac{x^3+y^3+z^3} {xyz}}+ \sqrt[2]{\frac{xy+yz+zx} {x^2 + y^2 + z^2}} \geq \sqrt[3]{3} + 1 [$]

### Re: inequality (a quickie)

Posted: September 22nd, 2019, 2:06 am
I think you need to transform LHS according to simple multiplication formulas and next use the fact that the arithmetic average is not smaller than the geometric one. I'd calculate it, but I got high on frankincense (https://www.mei.edu/sqcc/frankincense). Burnt sage is also good - to repel evil spirits (https://www.ncbi.nlm.nih.gov/pubmed/17030480)

### Re: inequality (a quickie)

Posted: September 22nd, 2019, 10:55 am
I found the spec. from LinkedIn. My solution was to break it into 2 independent optimiation problems,. compute gradient and set gradient to zero ==> x = y = z  = 0

For the 2nd term on the left-hand if you compute the gradient vector to zero you get a linear system giving  x = y = z for the minimum.

I did the gradient trick for the 1st terms as well and again x = y = z is the minimum. And if you plug in these values then equality is achieved.

// I have some frankinscense that I once bought near Sheeba's palace on the Yemen border along the Silk Road. Most beautiful and barren place It was 45 degrees...

### Re: inequality (a quickie)

Posted: September 22nd, 2019, 4:32 pm
We can split the problem. The firs part, just use the arithmetic and geomtric mean inequality ([$] (x_1 + x_2 + x_3) \geq 3 \sqrt[3] {x_1 x_2 x_3} [$])
with [$] x_1 = \frac{x^2}{y z}, x_2= \frac{y^2}{x z}, x_3 = \frac{x^2}{x y} [$] then use the cubic root of the inequality to obtain the first part.

For the second part, find the minimum of the (positive) function [$] f(x, y, z) = \frac{x^2 + y^2 + z^2}{x y + y z + z x} [$] by zeroing the gradient.
We find the solution as [$] (x, y, z) = t (1, 1, 1) [$] with a minimum 1.

### Re: inequality (a quickie)

Posted: September 22nd, 2019, 4:53 pm
I agree, part 1 is nice; you can also solve the first part by gradient as well.

There are an infinite number of solutions, yes? BTW you inverted [$]f(x,y,z)[$], what's the rationale?

### Re: inequality (a quickie)

Posted: September 22nd, 2019, 5:15 pm
Computing the gradient find the minimum or the maximum of the function. We are looking for a minimum (of a positive function).

### Re: inequality (a quickie)

Posted: September 22nd, 2019, 6:18 pm
?