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July 31[edit]

how to prove inequality plz give me hint[edit]

(x + y)(x + z) + (x + y)(y + z) + (x + z)(y + z) + x(x + 1)(y + z) + y(y + 1)(x + z) + z(z + 1)(x + y)≤ 2(n−3)(n-3)(n+6)/9, where n=x+y+z+3. — Preceding unsigned comment added by 39.35.233.28 (talk • contribs)

We don't do your homework for you. —moonythedwarf 17:21, 31 July 2021 (UTC)[reply]

I only want some hint

Is this true? Maybe I made a calculation error, but when I put x = y = 2 and z = −4, I get 12 for the left-hand side and 0 for the right-hand side. --Lambiam 18:26, 31 July 2021 (UTC)[reply]

Another counterexample is obtained for x = y = 6 and z = 1. So it is not a matter of the number tacitly being assumed positive. --Lambiam 19:08, 31 July 2021 (UTC)[reply]

o.k it is wrong, I found it in a research article, then can any body suggest a method to find an upper bond, provided that all these numbers are positive. Is it holds if x,y,z ≥ 3 or there is a counter example.

What is the research article in which you found it? --JBL (talk) 01:25, 1 August 2021 (UTC)[reply]

This looks like it could be handled with Lagrange multipliers. I get the bound given by plugging in x=y=z, which is certainly one of the critical points. But since this is a cubic function there are more than one critical point and the algebra gets rather hairy when you try to locate them. I think the upshot is that there are four critical points, one of which is a local extremum and the others are saddle points, which means analysis is needed on the boundary of the region. But it's not really clear what the region or its boundary are. A quadratic function would be much easier to handle, not even requiring calculus, since it can be written in terms of (x+y+z) and (x²+y²+z²-xy-xz-yz) ≥ 0. --RDBury (talk) 15:14, 1 August 2021 (UTC)[reply]

The inequality fails when x = y ≥ z+5. Are you looking for an upper bound having specifically the form f(x+y+z)? --Lambiam 20:25, 1 August 2021 (UTC)[reply]