Y by DrMath, jh235, TheCrafter, Eugenis, mathleticguyyy, somebodyyouusedtoknow, Adventure10, Mango247, NicoN9, cubres, ehuseyinyigit
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
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USA 97 [1/(b^3+c^3+abc) + ... >= 1/(abc)]
Maverick 48
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by SomeonecoolLovesMaths
Source: USAMO 1997/5; also: ineq E2.37 in Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu
Prove that, for all positive real numbers 
, 
, 
, the inequality
![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](//latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
holds.
, 
, 
, the inequality![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
holds.
48 replies
Hi inequalities
bjump 4
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by SomeonecoolLovesMaths
Source: OTIS, Kyle Wu
For positive real numbers 
, 
, and 
with 
prove that
, 
, and 
with 
prove that
4 replies
No more topics!
Advanced topics in Inequalities
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Y by Adventure10
Will edit later maybe
4.1:
5.1:
[/hide]
4.1:
by AM-GM so it suffices to show
but clearly 
giving the result.5.1:
so then
so it suffices to show
or 
.![$xy+yz+zx \ge 3 \sqrt[3]{x^2y^2z^2} \ge 3$](http://latex.artofproblemsolving.com/9/1/1/911e8dbf8e67fc68c37c72b890036797739666cb.png)
and
holds because 
h[/hide]
This post has been edited 2 times. Last edited by tastymath75025, Mar 7, 2015, 5:31 AM
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,
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![$x+y+z \geq 3\sqrt[3]{xyz} \geq 3$](http://latex.artofproblemsolving.com/3/1/4/314f429df203762a8392644a33d2f0d55cdd43d9.png)
and ![$xy+yz+xz \geq 3\sqrt[3]{x^2y^2z^2} \geq 3$](http://latex.artofproblemsolving.com/d/b/3/db32383ff1c260e0c691e722f49e7e02e0af9499.png)
.
,
.
and symmetrically others.
.
,
due to Cauchy,problem is over.
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Y by Strangett
Problem 5.1. (Tuan Le)
Let
be positive real numbers such that 
Prove the inequality
Problem4.2. (T.Q.Anh)
Let
and 
.Prove that
Problem 3.1. (Titu Andresscu)
Let
and 
Prove that
https://artofproblemsolving.com/community/c6h1838005p12333824
Let
be positive real numbers such that 
Prove the inequality
Problem4.2. (T.Q.Anh)
Let
and 
.Prove that
Problem 3.1. (Titu Andresscu)
Let
and 
Prove that
https://artofproblemsolving.com/community/c6h1838005p12333824This post has been edited 2 times. Last edited by sqing, Apr 16, 2025, 2:33 PM
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#6
Apr 16, 2025, 1:48 PM
Y by
You see NOTHING
This post has been edited 1 time. Last edited by giangtruong13, Apr 16, 2025, 1:55 PM
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#7
Apr 16, 2025, 2:54 PM
Y by
The direction of 3.2 is false.Anyway,here is an approach from "Problems from the book,Titu Andreescu and Gabriel Dospinescu"
and 
.
Proof:Apply Titu's Lemma and it succifies to show that 
or 
Unfortunaly,there is just a note which says "This is not an easy inequality.We will use the proposed problem 6 form the chepter "Two useful Substitutions" to reduce the above inequality to the form 
which follows from 
.
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are reals 
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Y by
Let 
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 3:27 PM
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.![\[
f''(x) = \frac{d^2}{dx^2} \ln(2x^2 + 3) = \frac{12}{(2x^2 + 3)^2} > 0 \quad \text{for all } x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/e/5/d/e5d9804fd04315833f902731f3cd4baaf563c7eb.png)
Hence, 
is strictly convex.![\[
f(a) + f(b) + f(c) \ge 3f\left( \frac{a + b + c}{3} \right) = 3f(1).
\]](http://latex.artofproblemsolving.com/9/0/3/903988a564a1e33e476d7a575cf090c3b13247eb.png)
Now,![\[
f(1) = \ln(2 \cdot 1^2 + 3) = \ln 5 \Rightarrow f(a) + f(b) + f(c) \ge 3 \ln 5 = \ln 125.
\]](http://latex.artofproblemsolving.com/3/1/9/3192bb0efde4cf7d3ffd7c2b5e0a3359e34d1783.png)
Exponentiating both sides gives:
.
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Y by
Novmath wrote:
Strangett wrote:
Problem 3.4. Given that 
, prove that
![\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]](//latex.artofproblemsolving.com/c/1/e/c1e561355e2a6d57be0f874a515ad8c4db9bb352.png)
, prove that![\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]](http://latex.artofproblemsolving.com/c/1/e/c1e561355e2a6d57be0f874a515ad8c4db9bb352.png)
Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?
Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
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Show that
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Y by
Strangett wrote:
Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
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Y by
sqing wrote:
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
The problem 2.1 (USAMO 2001 P3) is wrong. The condition was
Let
and satisfy 
Show that![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
Among a, b, c there must exist 2 numbers, say a and b, such that
Then:
On the other hand, from
we have:
Thus:
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Y by
sqing wrote:
Can you share solution of this one:
Problem: Prove that for all real numbers
, the following inequality holds:![\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]](http://latex.artofproblemsolving.com/d/a/7/da7212d7aab5c44dd78b993b3afa5dd938d50c6f.png)
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Y by
Novmath wrote:
sqing wrote:
Can you share solution of this one:
Problem: Prove that for all real numbers
, the following inequality holds:https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
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Y by
sqing wrote:
Novmath wrote:
sqing wrote:
Can you share solution of this one:
Problem: Prove that for all real numbers
, the following inequality holds:https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
Thank you very much
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Y by
sqing wrote:
Novmath wrote:
sqing wrote:
Can you share solution of this one:
Problem: Prove that for all real numbers
, the following inequality holds:https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
Bro where can I learn separation trick in inequalities ,do you have some source which explaines it detailed, if you send it to me it will be appreciated!
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and![\[
x = a + \frac{1}{b}, \quad y = b + \frac{1}{c}, \quad z = c + \frac{1}{a},
\]](http://latex.artofproblemsolving.com/e/f/e/efee2962579a7343e83a3e1b8e9f03af08b832bb.png)
then![\[
xy + yz + zx \geq 2 + x + y + z.
\]](http://latex.artofproblemsolving.com/7/e/7/7e7499966a274bdbf64234b7eef6a5093afa148f.png)
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#24
Apr 21, 2025, 8:22 PM
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or 
,This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 21, 2025, 8:34 PM
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Y by
sqing wrote:
Novmath wrote:
sqing wrote:
Can you share solution of this one:
Problem: Prove that for all real numbers
, the following inequality holds:https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
Please share the solution of Vasc one that I shared
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