Waste less time on Facebook — follow Brilliant.
×

Please Help! I'm Getting Stuck in Inequality

This inequality problem is from Phan Boi Chau High School for the Gifted Grade 10 Selection Test, Nghe An, Vietnam (Yes I'm very bad at inequalities):

Let \(a,b,c\) be positive reals. Find the minimum value of \(P=\dfrac{a^2}{(a+b)^2}+\dfrac{b^2}{(b+c)^2}+\dfrac{c}{4a}\).


I've done \(\dfrac{c}{4a}=\dfrac{c^2}{4ac}\geq\dfrac{c^2}{(c+a)^2}\).

I think the minimum value is \(\dfrac34\), so the problem is to prove \(\displaystyle \sum_\text{cyc}\dfrac{a^2}{(a+b)^2}\geq\dfrac34\). But I still can't.

Can you help?

Note by Tran Quoc Dat
1 year, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Continue the solution of @Ameya Daigavane . Using Cauchy-Schwarz \[\left(\frac{1}{1 + x}\right)^2 + \left(\frac{1}{1 + y}\right)^2 + \left(\frac{1}{1 + z}\right)^2 \geq\frac{1}{3}\bigg(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\bigg)^2\] So now we need to have \(f(x,y,z)=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq\frac{3}{2}\). Without losing generality, we assume that \(x\geq y\geq z\), then \(f(x,y,z)\geq f(x,y,y)\).

Now we consider \(f(x,y,y)\). Since \(x\geq y\) and \(xy^2=1\) we can have \(y\leq 1\) and \(x=\frac{1}{y^2}\). \[f(x,y,y)=\frac{1}{x+1}+\frac{2}{y+1}=\frac{y^2}{y^2+1}+\frac{2}{y+1}=\frac{y^3+3y^2+2}{(y^2+1)(y+1)}\] Now we prove \(f(x,y,y)\geq\frac{3}{2}\) \[\frac{y^3+3y^2+2}{(y^2+1)(y+1)}\geq\frac{3}{2}\] \[\Leftrightarrow\frac{2y^3+6y^2+4-3(y^2+1)(y+1)}{2(y+1)(y^2+1)}\geq 0\] \[\Leftrightarrow\frac{(1-y)^3}{2(y+1)(y^2+1)}\geq 0 \ (True \ since \ y\leq 1)\] So \(f(x,y,y)\geq\frac{3}{2}\Rightarrow f(x,y,z)\geq\frac{3}{2}\), therefore \(P\geq\frac{3}{4}\). The equality holds when \(x=y=z=1\) or \(a=b=c\). @Tran Quoc Dat , you can check out this solution, I don't know whether it's ok for grade 9 mathematically specialized students to apply this method

Gurīdo Cuong - 1 year, 5 months ago

Log in to reply

univariate solution always cool

Ms Ht - 1 year, 5 months ago

Log in to reply

Is this "mixing of variables"? I think there's some calculus theory behind this method.
Any non calculus approach?

Ameya Daigavane - 1 year, 5 months ago

Log in to reply

Did you mean the \(f(x, y, z) \geq f(x, y, y)\)?

Deeparaj Bhat - 1 year, 5 months ago

Log in to reply

@Deeparaj Bhat Oh no, he changed it, it was \(f(x,y,z) \geq f(x, \sqrt{yz}, \sqrt{yz}) \) I think.

Ameya Daigavane - 1 year, 5 months ago

Log in to reply

@Ameya Daigavane Ok. I guess the solution now doesn't use any calculus.

Deeparaj Bhat - 1 year, 5 months ago

Log in to reply

@Ameya Daigavane I've to change because later on, I found out that actually, \(f(x,y,z)\leq f(x,\sqrt{yz},\sqrt{yz})\)

Gurīdo Cuong - 1 year, 5 months ago

Log in to reply

I've tried but looks like there's no non calculus solution to this

Gurīdo Cuong - 1 year, 5 months ago

Log in to reply

After setting, \[x = \frac{b}{a}, y = \frac{c}{b}, z = \frac{a}{c} \] this becomes:
For \(x, y, z \in R_+\) such that \(xyz = 1\), \[ \left(\frac{1}{1 + x}\right)^2 + \left(\frac{1}{1 + y}\right)^2 + \left(\frac{1}{1 + z}\right)^2 \geq \frac{3}{4} \]

I believe Lagrange Multipliers is the best way here. I'm trying to find a non-calculus approach, and I've posted the question on AoPS as well.

Ameya Daigavane - 1 year, 5 months ago

Log in to reply

This Inequality was used in this problem,Viet Nam TST 2006 inequality problem: PROBLEM

Ms Ht - 1 year, 5 months ago

Log in to reply

Will Nesbitt's ineq work here?

Steven Jim - 6 months, 1 week ago

Log in to reply

It's nearly to Olympic more than selection test for grade 10.Huhmmmm

Ms Ht - 1 year, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...