Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with cauchy-schwarz-inequality
Search options answers only
not deleted
user 14830
The Cauchy-Schwarz inequality states $|\langle x,y \rangle |\leq ||x||\cdot ||y||.$ Use this tag for questions related to the CS inequality and its applications.
9
votes
How to prove that $1/ ((y+z) x^4) + 1/ ((z+x) y^4) + 1/ ((x+y) z^4) \geq 3/2$ for $x, y, z>0...
Is the "cauchy-schwarz-inequality" tag a guess or a hint? . . .
At any rate, it turns out to be a good start. Let
$$
R := \frac1{(y+z) x^4} + \frac1{(z+x) y^4} + \frac1{(x+y) z^4}.
$$
We show $xyz = …
- 79.9k