Find the minimum

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Find the minimum

G H J

Reveal topic tags

inequalities proposed

L

G H BBookmark kLocked kLocked NReply

Source: 2019 China Mathematical Olympiad Hope League Summer Camp

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#1 h Aug 10, 2019, 2:13 PM • 1 Y

Y by Adventure10

Let $x,y,z$ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+2(y+z)^2+3(z+x)^2.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#2 h Aug 10, 2019, 7:43 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $x,y,z$ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+2(y+z)^2+3(z+x)^2.$

Let $x,y,z$ be positive real number such that $xyz(x+y+z)=4.$ Prove that $(x+y)^2+2(y+z)^2+3(z+x)^2\geq 8\sqrt{11}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

906 posts

#3 h Aug 11, 2019, 3:41 PM • 4 Y

Y by teomihai, zhangruichong, Adventure10, Mango247

Denote $x+y=c; y+z=a; z+x=b$ .
Then $a,b,c$ are the lengths of the sides of a triangle (it holds $a+b>c$ and the similar relations).
Denote $p=\dfrac{a+b+c}{2}$ the semi-perimeter of the triangle.
$x=\dfrac{b+c-a}{2}=p-a$ and similarly $y=p-b; z=p-c; x+y+z=p$ .
Using the Heron's formula:
$xyz(x+y+z)=p(p-a)(p-b)(p-c)=S^2$ , where $S$ is the area of the triangle with the sides $a,b,c$ .
Results: $S=2$ .
If the altitude of the triangle, relative to the side $a$ is $h$ , we can write
$S=\dfrac{ah}{2}\Longrightarrow h=\dfrac{2S}{a}=\dfrac{4}{a}$ .

Consider the triangle $ABC$ , where $BC=a; AC=b; AB=c$ and the altitude $AD=h; AD\perp BC; D\in BC$ .
Denote $t=BD$ .
$(x+y)^2+2(y+z)^2+3(z+x)^2=2a^2+3b^2+c^2=2a^2+3[(a-t)^2+h^2]+(t^2+h^2)=$
$=4t^2-6at+5a^2+4h^2$ .
The second degree polynomial in the variable $t$ , $P(t)=4t^2-6at+5a^2+4h^2$ has the minimum for
$t_0=\dfrac{3a}{4}$ .
Results:
$(x+y)^2+2(y+z)^2+3(z+x)^2\ge P(t_0)=\dfrac{9a^2}{4}-\dfrac{18a^2}{4}+5a^2+\dfrac{64}{a^2}=\dfrac{11a^2}{4}+\dfrac{64}{a^2}$ .
Using $AM-GM$ ,
$\dfrac{11a^2}{4}+\dfrac{64}{a^2}\ge 2\cdot\sqrt{\dfrac{11a^2}{4}\cdot\dfrac{64}{a^2}}=8\sqrt{11}$ ,
hence $(x+y)^2+2(y+z)^2+3(z+x)^2\ge 8\sqrt{11}$ .

This post has been edited 1 time. Last edited by WallyWalrus, Aug 11, 2019, 3:54 PM
Reason: Expressions of a and b replaced

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

655 posts

#4 h Aug 11, 2019, 5:06 PM • 3 Y

Y by zhangruichong, Adventure10, Mango247

More generally we have the following inequality:
If $u,v,w\ge 0$ in any triangle $ABC$ we have
$ua^2+vb^2+wc^2\ge 4S\sqrt{uv+vw+wu}.$ Proof
so, if $u,v,w\ge 0$ then
$u(y+z)^2+v(z+x)^2+w(x+y)^2\ge 4\sqrt{xyz(x+y+z)(uv+vw+wu)}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#5 h Aug 11, 2019, 9:25 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive real number such that $abc(a+b+c)=4.$ Prove that
$(a+b)^2+(b+c)^2+ k(c+a)^2\geq 8\sqrt{2k+1}$ Where $k\in N^+.$

Let $a,b,c$ be positive real number such that $abc(a+b+c)=4.$ Prove that
$(a+b)^2+(b+c)^2+ (c+a)^2\geq 8\sqrt3$ $(a+b)^2+(b+c)^2+ 2(c+a)^2\geq 8\sqrt5$ $(a+b)^2+(b+c)^2+ 3(c+a)^2\geq 8\sqrt 7$ $(a+b)^2+(b+c)^2+ 4(c+a)^2 \geq 24$ $(a+b)^2+(b+c)^2+ 5(c+a)^2\geq 8\sqrt {11}$ $(a+b)^2+2(b+c)^2+ 2(c+a)^2\geq 16\sqrt 2$ $(a+b)^2+3(b+c)^2+3(c+a)^2\geq 8\sqrt {15}$ $(a+b)^2+4(b+c)^2+ 4(c+a)^2\geq 16\sqrt 6$ $(a+b)^2+5(b+c)^2+5(c+a)^2\geq 8\sqrt {35}$ $(a+1)^2+(b+1)^2+ (c+1)^2 >12$ Let $a,b$ be positive real number such that $ab(a+b)=1.$ Prove that $(a+1)^2+ (b+1)^2\geq 2+\sqrt[3]{2}+2\sqrt[3]{4}$ $(a+1)^2+2(b+1)^2>9$ k=1

This post has been edited 5 times. Last edited by sqing, Aug 10, 2021, 4:36 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#6 h Jan 23, 2021, 12:11 AM

Y by

Let $x,y,z$ be positive real number such that $xyz(x+y+z)=4.$ Prove that $(x+y)^2+(y+z)^2+(z+x)^2\geq 8\sqrt{3}.$ Solution: $(x+y)^2+(y+z)^2+(z+x)^2\geq 4(xy+yz+zx)\geq 4\sqrt{3xyz(x+y+z)}=8\sqrt{3}.$ Equality holds when $x=y=z=\sqrt[4]{\frac{4}{3}}.$

This post has been edited 2 times. Last edited by sqing, Jan 23, 2021, 3:02 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#7 h Mar 20, 2025, 8:26 AM

Y by

sqing wrote:

Let $a,b,c$ be positive real number such that $abc(a+b+c)=4.$ Prove that
$(a+b)^2+(b+c)^2+ k(c+a)^2\geq 8\sqrt{2k+1}$ Where $k\in N^+.$

Let $a,b,c$ be positive real number such that $abc(a+b+c)=4.$ Prove that
$(a+b)^2+(b+c)^2+ (c+a)^2\geq 8\sqrt3$

h
2020 Jilin High School League Preliminary:
Let $x,y,z$ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+(y+z)^2+(z+x)^2.$

Attachments:

This post has been edited 3 times. Last edited by sqing, Mar 20, 2025, 9:04 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a