Inequality involving x, y and z

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cefer

293 posts

cefer

#1 h Apr 28, 2012, 9:15 AM • 2 Y

Y by Adventure10, Mango247

Prove that
$\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),$
for all positive real numbers $x,y$ and $z$ .

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caoquyetthang

89 posts

caoquyetthang

#2 h Apr 28, 2012, 9:50 AM • 10 Y

Y by younesmath2012maroc, kgo, arandomperson123, TheImaginaryOne, hakN, ehuseyinyigit, Adventure10, Mango247, kiyoras_2001, and 1 other user

cefer wrote:

Prove that
$\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),$
for all positive real numbers $x,y$ and $z$ .

$LHS \ge \sum_{cyc}(x+y)(z+\sqrt{xy}) =$ $2\sum{xy}+\sum_{sym}(x+y)\sqrt{xy} \ge$ $2\sum{xy}+\sum_{sym}2\sqrt{xy}\sqrt{xy}=$ $4\sum{xy}=RHS$ .

[Moderator says: do not quote the entire post before yours]

This post has been edited 1 time. Last edited by Kunihiko_Chikaya, Jul 24, 2015, 4:29 AM

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delegat

652 posts

delegat

#3 h Apr 28, 2012, 9:52 AM • 2 Y

Y by Adventure10, Nari_Tom

From Cauchy-Schwartz-Bunyakovski inequality we get $\sqrt{\left(z+x\right)\left(z+y\right)}\geq z+\sqrt{xy}$
Then, using this one and $\left(x+y\right)\sqrt{xy}\geq2\sqrt{xy}\sqrt{xy}=2xy$
and summing similar inequalities for other 2 summands we get desired inequality.
(Seems to be very simple or I made some stupid mistake ???)

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delegat

652 posts

delegat

#4 h Apr 28, 2012, 9:55 AM • 2 Y

Y by caoquyetthang, Adventure10

caoquyetthang our solutions are actually same. Too easy question for such competition.
I forgot to add that $x=y=z$ is equality case, but it is obvious.

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Lawasu

212 posts

Lawasu

#5 h Apr 28, 2012, 11:03 AM • 10 Y

Y by PhantomR, Mathematicalx, kgo, TheImaginaryOne, Vladimir_Djurica, Adventure10, and 4 other users

Just take $x+y=a^2,\ y+z=b^2,\ z+x=c^2$ (so we will have $2x=a^2+c^2-b^2$ and the analogues one) and the inequality rewrites as:
$abc(a+b+c)\ge \displaystyle\sum(a^2+b^2-c^2)(a^2-b^2+c^2)\Leftrightarrow a^4+b^4+c^4+abc(a+b+c)\ge 2(a^2b^2+b^2c^2+c^2a^2)$
By Schur's Ineq for $r=2$ , we'll get $a^4+b^4+c^4+abc(a+b+c)\ge \sum\limits_{cyc}a^3b+\sum\limits_{cyc}ab^3\ge 2(a^2b^2+b^2c^2+c^2a^2)$ .

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dr_Civot

354 posts

dr_Civot

#6 h Apr 28, 2012, 11:10 AM • 4 Y

Y by hoangpham, Adventure10, Mango247, and 1 other user

cefer wrote:

Prove that
$\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),$
for all positive real numbers $x,y$ and $z$ .

First note that $(x+y)(y+z)=xy+yz+zx+y^2$ , so:
$3(xy+yz+zx)+(x^2+y^2+z^2)=(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)$
Now rewrite given inequality in:
$\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}\geq\sum_{cyc}(z+x)(z+y)+\sum_{cyc}xy-\sum_{cyc}x^2$
Let $x+y=a,y+z=b,z+x=c$ . Note that $a,b,c$ are sides of a triangle and note that:
$2\sum_{cyc}x^2-2\sum_{cyc}xy=\sum_{cyc}\left((x+z)-(y+z)\right)^2$
$2\sum_{cyc}(z+x)(z+y)-2\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}=\sum_{cyc}\left(\sqrt{x+z}-\sqrt{y+z}\right)(x+y)$
Now rewrite inequality in:
$\sum_{cyc}\left(\sqrt{a}-\sqrt{b}\right)^2\left(a+b-c+2\sqrt{ab}\right)\geq 0$
Which is true since $a+b>c.$ $\square$

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vanthanhlhp

215 posts

vanthanhlhp

#7 h Apr 28, 2012, 11:54 AM • 2 Y

Y by Adventure10 and 1 other user

Rewrite the inequality as
$\sum {\frac{1}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }}} \ge \frac{{4\left( {xy + yz + zx} \right)}}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}$
Using the familiar inequality $\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) \ge \frac{8}{9}\left( {xy + yz + zx} \right)\left( {x + y + z} \right)$ , it reduces to prove
$\sum {\frac{1}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }}} \ge \frac{9}{{2\left( {x + y + z} \right)}}$
Applying Cauchy and Cauchy Schwarz inequality we get
$\sum {\frac{1}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }}} \ge \sum {\frac{2}{{x + y + 2z}}} \ge \frac{9}{{2\left( {x + y + z} \right)}}$
Complete proof. Equality holds when $x=y=z$ .

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teps

131 posts

teps

#8 h Apr 28, 2012, 2:08 PM • 6 Y

Y by hctb00, MathsLion, Adventure10, Mango247, and 2 other users

Also $A\ge G$ on the whole expression + $9(x+y)(y+z)(x+z)\ge 8(x+y+z)(xy+yz+xz)$ does the trick.

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enndb0x

843 posts

enndb0x

#9 h Apr 28, 2012, 8:45 PM • 3 Y

Y by leozitz, Adventure10, Mango247

Method 1. Use Cauchy Schwarz to get $\sqrt{(z+x)(z+y)}\geqslant z+\sqrt{xy}$ . Hence $\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geqslant \sum_{cyc}(x+y)(z+\sqrt{xy}) =\sum_{cyc}z(x+y) +\sum_{cyc}(x+y)\sqrt{xy}$ .
Notice that by AM-GM $x+y \geqslant 2\sqrt{xy}$ , which gives $\sum_{cyc}(x+y)\sqrt{xy} \geqslant \sum_{cyc}2xy$ . Therefore
$LHS \geqslant \sum_{cyc}z(x+y) +\sum_{cyc}(x+y)\sqrt{xy} \geqslant 2(xy+yz+zx) +2(xy+yz+zx)=4(xy+yz+zx)$

Method 2. Without loss of generality, let $xy+yz+zx=1$ . Then we have to prove that
$\sum_{cyc}(x+y)\sqrt{z^2 +1} \geqslant 4$
Observe that $f(x)=\sqrt{x^2 +1}$ is a convex function of $x$ . We proceed by Jensen's inequality:

$2(x+y+z)\sum_{cyc}\frac{x+y}{2(x+y+z)} \sqrt{z^2 +1} \geqslant 2(x+y+z) \sqrt{\frac{1}{(x+y+z)^2}+1}$
Set $p=x+y+z$ . We want to prove that $2p\sqrt{\frac{1+p^2}{p^2}}= 2\sqrt{1+p^2} \geqslant 4$ .

By AM-GM $p^2 =(x+y+z)^2 \geqslant 3(xy+yz+zx)=3$ , hence $2\sqrt{1+p^2} \geqslant 2\cdot 2 =4$ . Done.

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Abu3gab

8 posts

Abu3gab

#10 h Apr 29, 2012, 1:33 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$\begin{gathered} (x - y)^2 \geqslant 0 \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + y^2 ) \geqslant (xy + yz + zx)2xy \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + y^2 ) + (xy + yz + zx)^2 + x^2 y^2 \geqslant \hfill \\ (xy + yz + zx)2xy + (xy + yz + zx)^2 + x^2 y^2 \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + y^2 ) + (xy + yz + zx)^2 + x^2 y^2 \geqslant \hfill \\ (xy + (xy + yz + zx))^2 \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + xy + yz + zx) + y^2 (x^2 + xy + yz + zx) \geqslant \hfill \\ (2xy + yz + zx)^2 \hfill \\ \hfill \\ \Leftrightarrow (x^2 + xy + yz + zx)(y^2 + xy + yz + zx) \geqslant \hfill \\ (2xy + yz + zx)^2 \hfill \\ \hfill \\ \Leftrightarrow (x + y)(x + z)(x + y)(y + z) \geqslant (2xy + yz + zx)^2 \hfill \\ \hfill \\ \Leftrightarrow (x + y)\sqrt {(x + z)(y + z)} \geqslant (2xy + yz + zx) \hfill \\ \hfill \\ \Rightarrow \sum\limits_{cyc} {(x + y)\sqrt {(x + z)(y + z)} } \geqslant 4(xy + yz + zx) \hfill \\ \end{gathered}$

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Elio (n)

569 posts

Elio (n)

#11 h Apr 29, 2012, 2:23 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $a,b,c$ be the sides of a triangle such that $x=b+c-a, y=a+c-b, z=a+b-c$ then we will have : $\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+a^{2}+b^{2}+c^{2}\ge 2(ab+bc+ac)$ which is equivalent with :

$(a-b)^{2}+(b-c)^{2}+(a-c)^{2}-a(\sqrt{c}-\sqrt{b})^{2}+b(\sqrt{a}-\sqrt{c})^{2})+c(\sqrt{a}-\sqrt{b})^{2}\ge 0$ factorizing we get the following which is true by triangle inequality:

$\sum{(\sqrt{a}-\sqrt{b})^{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})}\ge0$

If $a,b,c$ can be the sides of a triangle then $\sqrt{a},\sqrt{b},\sqrt{c}$ can also be triangle sides

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WakeUp

1347 posts

WakeUp

#12 h Apr 29, 2012, 5:24 PM • 2 Y

Y by Adventure10 and 1 other user

The problem is equivalent to a problem of the Romanian TST 2001:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2148433

The very form of the problem statement is actually given in Akiyama's post..

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DHu

4 posts

DHu

#13 h May 2, 2012, 8:31 PM • 3 Y

Y by GioOrnikapa, Adventure10, and 1 other user

This suddenly came to me at midnight on the competition day (I got 0 in the exam): Do normal substitutions of a = root(y+z) and so on and prove that a,b,c satisfy triangle inequality, and then $LHS=abc(a+b+c)=2abc\times s$ $RHS=16\times Area^2$ where s is the semi-perimeter. But this easily cancels using well known formulae for the Area of triangle made by sides a,b,c. $RHS=16\times \frac{abc}{4R} \times rs$
This cancels down nicely to Euler's inequality: $R\geq 2r$

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frill

316 posts

frill

#14 h May 31, 2012, 6:40 PM • 1 Y

Y by Adventure10

$\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}\geq 4(xy+yz+zx),$

And sub in $a=\sqrt{x+y}\ , b=\sqrt{x+z}\ , c=\sqrt{y+z}$

Then you get $abc(a+b+c)\geq(a^{2}+b^{2}-c^{2})(a^{2}-b^{2}+c^{2})+(a^{2}-b^{2}+c^{2})(b^{2}+c^{2}-a^{2})+(b^{2}+c^{2}-a^{2})(a^{2}+b^{2}-c^{2})$

Note that the triangle inequality holds for a,b and c as a, b, c, x, y, z are all positive so $a^{2}+b^{2}>c^{2}\Rightarrow\ a+b>c$ and similarly for others.

Thus dividing by abc we get: $a+b+c\geq\ a\cos{B}\cos{C}+b\cos{C}\cos{A}+c\cos{A}\cos{B}$ which is trivial as cos(x) is at most 1 for all real x

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Markomak1

514 posts

Markomak1

#15 h May 31, 2012, 7:53 PM • 2 Y

Y by Adventure10, Mango247

Frill, when does equality hold at your inequality?

Also i think you need to prove that if with $a,b,c$ you can form a triangle, then you can also form a triangle with side lengths $\sqrt{a},\sqrt{b},\sqrt{c}$

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arqady

30152 posts

arqady

#34 h Mar 16, 2016, 8:09 AM • 1 Y

Y by Adventure10

luofangxiang wrote:

The following inequality is also true
Let $a,b,c\geq0$ , $\sum_{cyc}ab=3$ ; Prove: $\sum_{cyc}\sqrt{(a^2+3)(b^2+3)}\geq3\sum_{cyc}a+3$

We need to prove that $\sum_{cyc}(a+b)\sqrt{(a+c)(b+c)}\geq(a+b+c)\sqrt{3(ab+ac+bc)}+ab+ac+bc$ or
$\sum_{cyc}(2a^2+4ab)-2(a+b+c)\sqrt{3(ab+ac+bc)}\geq\sum_{cyc}(a+b)\left(a+b+2c-2\sqrt{(a+c)(b+c)}\right)$ or
$2(a+b+c)\left(a+b+c-\sqrt{3(ab+ac+bc)}\right)\geq\sum_{cyc}(a+b)\left(\sqrt{a+c}-\sqrt{b+c}\right)^2$ or
$\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}\right)\geq0$ , which is true because
$\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}=$
$=\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b}{a+b+2c+2\sqrt{(a+c)(b+c)}}\geq$
$\geq\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b+c}{a+b+c+2\sqrt{ab+ac+bc}}\geq0$ . Done!

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luofangxiang

4613 posts

luofangxiang

#35 h Mar 16, 2016, 8:12 AM • 1 Y

Y by Adventure10

Nice，arqady

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xzlbq

15849 posts

xzlbq

#37 h Dec 31, 2016, 6:33 AM • 2 Y

Y by Adventure10, Mango247

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }$ $\geq { \frac {288}{125}}\,{\frac { \left( yz+ \left( z+x \right) \left( x+y \right) \right) \left( zx+ \left( x+y \right) \left( y+z \right) \right) \left( xy+ \left( y+z \right) \left( z+x \right) \right) } { \left( y+z \right) \left( z+x \right) \left( x+y \right) \left( x +y+z \right) }}\geq 4(xy+yz+zx)$
${\frac {9}{16}}\,{\frac { \left( 2\,y+z+x \right) \left( 2\,z+x+y \right) \left( z+2\,x+y \right) }{x+y+z}}\geq \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }$

This post has been edited 3 times. Last edited by xzlbq, Dec 31, 2016, 6:41 AM

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xzlbq

15849 posts

xzlbq

#38 h Dec 31, 2016, 3:39 PM • 1 Y

Y by Adventure10

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-{ \frac {288}{125}}\, \left( {\frac {yz}{y+z}}+{\frac { \left( z+x \right) \left( x+y \right) }{y+z}} \right) \left( {\frac {zx}{z+x}} +{\frac { \left( x+y \right) \left( y+z \right) }{z+x}} \right) \left( {\frac {xy}{x+y}}+{\frac { \left( y+z \right) \left( z+x \right) }{x+y}} \right) \left( x+y+z \right) ^{-1}\geq 0.$

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-4 \,{\frac { \left( x+y+z \right) \left( x{y}^{2}+y{z}^{2}+{x}^{2}z \right) }{{x}^{2}+{y}^{2}+{z}^{2}}}\geq 0.$

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-6 \, \left( {\frac { \left( y+z \right) ^{2}}{2\,{x}^{2}+yz}}+{\frac { \left( z+x \right) ^{2}}{2\,{y}^{2}+zx}}+{\frac { \left( x+y \right) ^{2}}{2\,{z}^{2}+xy}}+2\,{\frac {xy+zx+yz}{{x}^{2}+{y}^{2}+{z}^{2}}} \right) xyz \left( x+y+z \right) ^{-1}\geq 0$

$- \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }- \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }- \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }+{ \frac {9}{16}}\,{\frac { \left( 2\,y+z+x \right) \left( 2\,z+x+y \right) \left( z+2\,x+y \right) }{x+y+z}}\geq 0$

$-12\,{\frac {\sqrt {xy}{z}^{2}}{x+y+z}}-12\,{\frac {\sqrt {yz}{x}^{2}} {x+y+z}}-12\,{\frac {\sqrt {zx}{y}^{2}}{x+y+z}}+ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }\geq 0$

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-3 \,{\frac {{x}^{2} \left( y+z \right) ^{3}}{ \left( x+y+z \right) \left( {y}^{2}+{z}^{2} \right) }}-3\,{\frac {{y}^{2} \left( z+x \right) ^{3}}{ \left( x+y+z \right) \left( {z}^{2}+{x}^{2} \right) } }-3\,{\frac {{z}^{2} \left( x+y \right) ^{3}}{ \left( x+y+z \right) \left( {x}^{2}+{y}^{2} \right) }}\geq 0$

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-6 \,{\frac {{x}^{2} \left( y+z \right) ^{3}}{ \left( z+x \right) \left( x+y+z \right) \left( x+y \right) }}-6\,{\frac {{y}^{2} \left( z+x \right) ^{3}}{ \left( x+y \right) \left( x+y+z \right) \left( y+z \right) }}-6\,{\frac {{z}^{2} \left( x+y \right) ^{3}}{ \left( y+z \right) \left( x+y+z \right) \left( z+x \right) }}\geq 0$

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-6 \,{\frac { \left( x+y \right) ^{2}xyz}{ \left( x+y+z \right) \left( { z}^{2}+xy \right) }}-6\,{\frac { \left( y+z \right) ^{2}xyz}{ \left( x +y+z \right) \left( {x}^{2}+yz \right) }}-6\,{\frac { \left( z+x \right) ^{2}xyz}{ \left( x+y+z \right) \left( {y}^{2}+zx \right) }}\geq 0$

$\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }- 12\,{\frac { \left( x+y \right) ^{2}xyz}{ \left( y+z \right) \left( x +y+z \right) \left( z+x \right) }}-12\,{\frac { \left( y+z \right) ^{ 2}xyz}{ \left( z+x \right) \left( x+y+z \right) \left( x+y \right) } }-12\,{\frac { \left( z+x \right) ^{2}xyz}{ \left( x+y \right) \left( x+y+z \right) \left( y+z \right) }}\geq 0.$

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M.Unubold

80 posts

M.Unubold

#39 h Mar 25, 2018, 9:16 AM • 2 Y

Y by Adventure10, Mango247

Correct my solution please?
Write given inequality $\sum_{cyc} \dfrac{x+y}{2(x+y+z)} \cdot \sqrt{ z^{2}+(xy+yz+zx)} \geq \dfrac{2(xy+yz+zx)}{x+y+z}$ .
Observe that function $f(t)=\sqrt {t^{2}+a}$ for fixed $a$ positive real number and $t$ being positive real, $f$ is convex.So, by assuming $xy+yz+zx=a$ is constant in above inequality, we have(by Jensen's inequality) $LHS= \sum_{cyc} \dfrac{x+y}{2(x+y+z)} \cdot f(z) \geq f \bigg(\sum_{cyc} \dfrac{z(x+y)}{2(x+y+z)} \bigg) =f \bigg(\dfrac{xy+yz+zx}{x+y+z} \bigg)=\sqrt {\dfrac{(xy+yz+zx)^{2}}{(x+y+z)^{2}}+( xy+yz+zx)}$ .Let $a=xy+yz+zx$ and $b=x+y+z$ .Then $\sqrt {\dfrac{(xy+yz+zx)^{2}}{(x+y+z)^{2}}+( xy+yz+zx)}=\sqrt{\dfrac{a^{2}}{b^{2}}+a}$ and so $\sqrt{\dfrac{a^{2}}{b^{2}}+a} \geq \dfrac{2(xy+yz+zx)}{x+y+z}=\dfrac{2a}{b}$ is equivalent to $b^{2} \geq 3a$ or $(x+y+z)^{2} \geq 3(xy+yz+zx)$ and latter inequality is true, since $(x+y+z)^{2}-3(xy+yz+zx)=\frac{1}{2} \cdot ((x-y)^{2}+(y-z)^{2}+(z-x)^{2}) \geq 0$

This post has been edited 3 times. Last edited by M.Unubold, Mar 25, 2018, 9:23 AM
Reason: typo

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M.Unubold

80 posts

M.Unubold

#40 h Mar 28, 2018, 1:31 AM • 2 Y

Y by Adventure10, Mango247

Correct solution this time.
Let $x+y=2c$ , $y+z=2a$ and $z+x=2b$ for positive $a,b$ and $c$ . We have $x=b+c-a$ , $y=c+a-b$ and $z=a+b-c$ . Therefore given inequality reduces to $\sum_{cyc} 2c \sqrt{4ab} \geq 4 \big(\sum_{cyc} (a+b-c)(b+c-a) \big)$ or $\sum_{cyc} c\sqrt{ab} \geq 2ab+2bc+2ca-a^{2}-b^{2}-c^{2}$ . Latter inequality is equivalent to $a^{2}+b^{2}+c^{2}-ab-bc-ca \geq ab+bc+ca-a\sqrt{bc}-b\sqrt{ca}-c\sqrt{ab}$ . Note that $x^{2}+y^{2}+z^{2}-xy-yz-zx=\frac{1}{2} \big((x-y)^{2}+(y-z)^{2}+(z-x)^{2} \big)$ . Using this identity for $(x,y,z)=(a,b,c) , (\sqrt{ab},\sqrt{bc},\sqrt{ca})$ , we have last inequality is equivalent to $\sum_{cyc} (a-b)^{2} \geq \sum_{cyc} (\sqrt{ca}-\sqrt{bc})^{2}$ . Now note that $(a-b)^{2}-(\sqrt{ca}-\sqrt{bc})^{2} =(\sqrt{a}-\sqrt{b})^{2} \cdot \big((\sqrt{a}+\sqrt{b})^{2}-c \big)=(\sqrt{a}-\sqrt{b})^{2} \cdot (a+b-c+2\sqrt{ab})$ and since $a+b-c=z >0$ , we have $(a-b)^{2} \geq (\sqrt{ca}-\sqrt{bc})^{2}$ so desired inequality proved

This post has been edited 3 times. Last edited by M.Unubold, Mar 28, 2018, 1:33 AM
Reason: typo

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TheImaginaryOne

109 posts

TheImaginaryOne

#41 h Mar 30, 2018, 3:30 AM • 1 Y

Y by Adventure10

We have $RHS=\sum_{cyc}2x(y+z)$ .
Let $a,b,c=x+y,y+z,z+x$
then the inequality is equivalent to
$\sum_{cyc} a\sqrt{bc} \ge \sum_{cyc}2\left(\frac{a+b-c}{2}\cdot c\right)$ $\sum_{cyc} \left(a\sqrt{bc}+c^2\right) \ge \sum_{cyc} 2ac$
To prove the last line, let $a,b,c = p^2,q^2,r^2$ . We use 4th degree Schur and then AM-GM:

$\sum_{cyc} \left(p^2qr + r^4\right) \ge \sum_{cyc} \left(p^3r + pr^3\right) \ge \sum_{cyc}2p^2r^2$ as required.

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Feridimo

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Feridimo

#42 h Jan 24, 2020, 8:29 AM • 1 Y

Y by Adventure10

$My$ $solution$
$z+x\geq 2\sqrt{zx},y+x\geq 2\sqrt{yx},x+y\geq 2\sqrt{xy}$
$Then$
$\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 2\sqrt{xy}\sqrt{2\sqrt{xz}2\sqrt{yz}}=4(xy+xz+yz)$

This post has been edited 2 times. Last edited by Feridimo, Jan 24, 2020, 8:30 AM

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dastan111

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dastan111

#43 h Jan 24, 2020, 12:53 PM • 1 Y

Y by Adventure10

$\left\{ \begin{gathered} x+y= c\\ y+z=a \\ z+x=b \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x=\frac{b+c-a}{2} \\ y=\frac{a+c-b}{2} \\ z=\frac{a+b-c}{2} \\ \end{gathered} \right.\Rightarrow$ $\Rightarrow c\sqrt{ab}+a\sqrt{bc}+b\sqrt{ac} \geq 2ab+2bc+2ac-(a^2+b^2+c^2) \Rightarrow$ $\Rightarrow \underbrace{2ab+2bc+2ac-(a^2+b^2+c^2)-\left(c\sqrt{ab}+a\sqrt{bc}+b\sqrt{ac} \right)}_S \leq 0$ $0 \geq S \geq 2ab+2bc+2ac-(a^2+b^2+c^2)-c\frac{a+b}{2}-a\frac{b+c}{2}-b\frac{a+c}{2}=$ $= ab+bc+ac-(a^2+b^2+c^2)=-(a^2+b^2+c^2-ab-bc-ac)=$ $=-\frac{1}{2} \left( (a-c)^2+(c-b)^2+(b-a)^2 \right)$

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agwwtl03

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agwwtl03

#44 h Jun 17, 2021, 9:48 PM • 1 Y

Y by Mango247

Claim: $(x+y)\sqrt{(z+x)(z+y)} \geq 2xy+yz+zx$
Proof: After we square both sides, it suffices to show $x^3y+xy^3+x^3z+y^3z \geq 2x^2y^2 +x^2yz+xy^2z$ . But by AM-GM, we have $x^3y+xy^3\geq 2x^2y^2$ and $x^3z+y^3z\geq x^2yz+xy^2z\square$

So, $\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq \sum_{cyc}2xy+yz+zx =4(xy+yz+zx)$ and we are done.

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MathThm

54 posts

MathThm

#46 h Aug 13, 2021, 7:39 AM

Y by

Solution :
Let $x+y=a^2,y+z=b^2$ and $z+x=c^2$ we can write the inequality as
$\sum_{cyc}a^2bc\geq 4\sum_{cyc}xy$ we can write this as
$\sum_{cyc}a^2bc+\sum_{cyc}a^4\geq 2\sum_{cyc}a^2b^2$ we can see the LHS as $\sum_{cyc}a^2(a^2+bc)$ which is basically schur's inequality so we can write this as
$\implies \sum_{cyc}a^2(a^2+bc)\geq \sum_{cyc}a^3(b+c)$ which follows by schur's inequality so its suffince to prove that
$\sum_{sym}a^3b\geq \sum_{cyc}a^2b^2$ which is true since $(3.1)\succ(2,2)$ which is true by muirhead and we are done $\blacksquare$

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john0512

4171 posts

john0512

#47 h Jan 7, 2023, 7:17 PM

Y by

me when

Make the substitution $a=x+y,b=y+z,c=z+x.$ It can then be seen that $x=\frac{a+c-b}{2}$ etcera. Therefore, we have $4(xy+xz+yz)=2(ab+ac+bc)-(a^2+b^2+c^2).$ Therefore, it remains to show that $a^2+b^2+c^2+a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 2(ab+ac+bc).$ We rewrite this as $cyc(2,0,0)+cyc(1,1/2,1/2)\geq 2cyc(1,1,0).$ We can double the degrees (essentially the substitution $a=a_2 ^2$ etc, so it becomes $cyc(4,0,0)+cyc(2,1,1)\geq 2cyc(2,2,0).$ By Schur with $r=2$ , it follows that $cyc(4,0,0)+cyc(2,1,1)\geq sym(3,1,0),$ and by Muirhead it then follows that $sym(3,1,0)\geq sym(2,2,0)=2cyc(2,2,0),$ so we are done.

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tesla501

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tesla501

#48 h Apr 24, 2023, 5:50 PM

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$\begin{align*} \sum_{cyc}^{} (x + y) \sqrt { (x + z)(y + z) } &= \sum_{cyc}^{} \sqrt{ (x + y)(x + z) } \cdot \sqrt{ (x + y)(y + z) } \\ \sum_{cyc}^{} \sqrt{ (x + y)(x + z) } \cdot \sqrt{ (x + y)(y + z) } &\geq \frac {2} { \frac{1} {(x + y)(x + z)} + \frac {1} {(x + y)(y + z)}} = \sum_{cyc}^{} \frac {2(x+y)(y+z)(z + x)} {(x + z) + (y + z)} = \\ 2(x + y)(y + z)(z + x) \sum_{cyc}^{} \frac {1} {(x + z) + (y + z)} &\text{ vs } 4(xy + yz + zx) \\ (x + y)(y + z)(z + x) \sum_{cyc}^{} \frac {1} {(x + z) + (y + z)} &\text{ vs } 2(xy + yz + zx) \end{align*}$ Notice that:
$\begin{align*} &(\sum_{cyc}^{} \frac {1} {(x + z) + (y + z)})( \sum_{cyc}^{} ((x + z) + (y + z)) \geq (1 + 1 + 1)^2 \\ &4(\sum_{cyc}^{} \frac {1} {(x + z) + (y + z)})(x + y + z) \geq 9 \\ &\sum_{cyc}^{} \frac {1} {(x + z) + (y + z)} \geq \frac{9} { 4(x + y + z) } \end{align*}$ So we have to prove:
$\begin{align*} \frac{9(x + y)(y + z)(z + x)} { 4(x + y + z) } &\geq 2(xy + yz + zx) \\ 9(x + y)(y + z)(z + x) &\geq 8 (x + y + z)( xy + yz + zx ) \\ 9 (\sum_{sym}^{} x^2y + 2xyz) &\geq 8 ( \sum_{sym}^{} + 3xyz) \\ 9 \sum_{sym}^{} x^2y + 18xyz &\geq 8 \sum_{sym}^{} x^2y + 24xyz \\ \sum_{sym}^{} x^2y &\geq 6xyz \\ x^2y + x^2z + y^2z + y^2x + z^2x + z^2y &\geq 6xyz \end{align*}$ Which is AM-GM.

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ehuseyinyigit

783 posts

ehuseyinyigit

#51 h Oct 21, 2023, 7:36 PM

Y by

Generalization 1
Let $a_{1},a_{2},\cdots,a_{n}(n\geq 3)$ be positive reels. Then prove that

$\sum_{cyc}{\left((a_{1}+a_{2}+\cdots+a_{n-1})\sqrt[n-1]{\prod_{i=1}^{n-1}{\left(a_{n}+a_{i}\right)}}\right)}\geq 2\left(\sum_{i=k+1-n}^{n-1}{a_{i}a_{k}}\right)+(n-1)\left(\sum_{cyc}{\left(a_{1}a_{2}\cdots a_{n-1}\right)^{\dfrac{n-1}{2}}}\right)$

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Baimukh

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Baimukh

#52 h Mar 20, 2025, 9:59 PM

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$(x+y)\sqrt {(z+x)(z+y)}+(y+z)\sqrt {(x+y)(x+z)}+(z+x)\sqrt {(y+z)(y+x)}\ge 3\sqrt[3]{(x+y)^2(y+z)^2(z+x)^2}\ge 3\sqrt[3]{\dfrac{64}{81}(x+y+z)^2(xy+yz+zx)^2}\ge 3\sqrt[3]{\dfrac{64}{27}(x+y+z)^3}=4(xy+yz+zx)$

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