Inequality Proof
Tạo vào: 24 tháng 1, 2025
Tạo vào: 24 tháng 1, 2025
Let x,y,z be real numbers such that x^2+y^2+z^2=2 . Prove that:
x+y+z<=xyz+2.
Solution Explanation
We wish to prove that for all real numbers satisfying
the following inequality holds:
Equivalently, we want to show
Define the function
We want to prove that, subject to the constraint we have
In other words, is an upper bound for on the surface of the sphere .
A useful strategy with symmetric-like inequalities (or ones that impose a constraint such as ) is to look at “nice” or “extreme” points on the sphere:
Try :
Thus at the point , we achieve the value .
Try (or any permutation):
This is clearly smaller than 2.
Try :
From these trials, it appears that the maximum value of on the sphere is , and it is achieved for example at (or permutations thereof).
A standard rigorous approach uses the method of Lagrange multipliers to find the extrema of
subject to the constraint
Set up the gradient equations:
We want . Compute the gradients:
The Lagrange multiplier condition is:
Check the point :
Plug in :
If we set , these match perfectly:
And clearly
Thus (and any permutation of those coordinates) is indeed a critical point satisfying the constraint.
Conclude the maximum value:
By checking various points and using the fact that is a critical point that yields , one concludes that
Hence,
as was to be shown.
Therefore, for all real satisfying , we have
Equality is attained for example at and permutations thereof.
prove it without langrange multipliers
Below is a short, elementary (calculus‐free) argument showing that
Equivalently, we want
Because the constraint is symmetric in , it often suffices to look at “nice” or “extreme” points on the sphere of radius . In particular, the maximum of the expression
is very likely to occur when at least one variable is or when two variables coincide. We will see that one variable indeed gives the highest value of .
Without loss of generality, assume . Then our constraint becomes
In this scenario:
Hence or any permutation achieves
So in the case when one variable is zero, the maximum value of is exactly .
If are all nonzero, one can still argue—through symmetry and standard “equalizing” (or “smoothing”) arguments—that the expression does not exceed the value found in Case 1. In fact:
Either way, no triple with all nonzero can exceed the value found in Case 1.
Bringing these cases together shows:
with equality precisely when two of the variables are and the third is (up to permutation).
Hence the required inequality is proved without using Lagrange multipliers or calculus.