Triangle Inequality.

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Source: Chinese Northern Mathematical Olympiad 2007

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1270 posts

April

#1 h Aug 5, 2007, 9:05 AM • 2 Y

Y by Adventure10, Mango247

Let $a,\, b,\, c$ be side lengths of a triangle and $a+b+c = 3$ . Find the minimum of
$a^{2}+b^{2}+c^{2}+\frac{4abc}{3}$

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91 posts

eLeM

#2 h Aug 5, 2007, 9:59 AM • 1 Y

Y by Adventure10

April wrote:

Let $a,\, b,\, c$ be side lengths of a triangle and $a+b+c = 3.$ Find the minimum of
$a^{2}+b^{2}+c^{2}+\frac{4abc}{3}$
P.S. Thank you, Zhaoli.

Since $a,b,c$ are sides of a triangle we can change $a=x+y$ , $b=y+z$ and $c=z+x$ with $x+y+z=\frac{3}{2}$
So we have:

$a^{2}+b^{2}+c^{2}+\frac{4abc}{3}=\frac{(a^{2}+b^{2}+c^{2})(a+b+c)+4abc}{3}=$ $\frac{2((x+y)^{2}+(y+z)^{2}+(z+x)^{2})(x+y+z)+4(x+y)(y+z)(z+x)}{3}=$ $\frac{4(x^{3}+y^{3}+z^{3}+3x^{2}y+3xy^{2}+3y^{2}z+3yz^{2}+3z^{2}x+3zx^{2}+5xyz)}{3}=$
$\frac{4((x+y+z)^{3}-xyz)}{3}=\frac{4(\frac{26}{27}(x+y+z)^{3}+(\frac{x+y+z}{3})^{3}-xyz)}{3}\geq$
$\geq \frac{4(\frac{26}{27}(x+y+z)^{3}}{3}=\frac{13}{3}$

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528 posts

#3 h Aug 6, 2007, 5:24 AM • 2 Y

Y by Adventure10, Mango247

the third degree Schur's inequality $(a+b+c)^{3}+9abc\geq\ 4(a+b+c)(ab+ac+bc)---\to abc\geq \frac{1}{9}(12(ab+ac+bc)-27)$ and $\frac{1}{3}(a+b+c)^{2}\geq\ ab+ac+bc$ from this
$a^{2}+b^{2}+c^{2}+\frac{4abc}{3}=(a+b+c)^{2}-2ab-2ac-2bc+\frac{4}{3}*\frac{1}{9}(12(ab+ac+bc)-27)=3^{2}-2(ab+ac+bc)+\frac{16}{9}(ab+ac+bc)-4=5-\frac{2}{9}(ab+ac+bc)\geq\ 5-\frac{6}{9}=\frac{13}{4}$
a=b=c=1

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679 posts

#4 h Aug 10, 2007, 8:43 AM • 1 Y

Y by Adventure10

Let x= $\frac{a}{3}$ , y= $\frac{b}{3}$ , z= $\frac{c}{3}$ so that $x+y+z =1.$

Now $a^{2}+b^{2}+c^{2}$ + $\frac{4ab}{3}$ = $(a+b+c)^{2}$ - 2 $\sum$ ab + $\frac{4ab}{3}$ which after substitution for a,b c, in terms of x,y,z and using a+b+c =3 becomes

$9-18(xy+yz+zx-2xyz)$ . We will find the maximum of $xy+yz+zx-2xyz.$

The identity $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$ , after replacing x+y+z by 1 and some re-arrangement gives

$xy+yz+zx-2xyz$ = $x^{2}(1-x)$ + $y^{2}(1-y)$ + $z^{2}(1-z)$ + $xyz$

Assume that x $\leq$ y $\leq$ z . Then 1-x $\geq$ 1-y $\geq$ 1-z
Writing $\sum$ $x^{2}(1-x)$ as $\sum(x.x.(1-x) )$ ,Tchebychefs inequality gives

$ \frac{\sum(x.x.(1-x)}_{3}$ $\leq$ $ \frac{\sum(x)}_{3}$.$ \frac{\sum(x)}_{3}$.$ \frac{\sum(1-x)}_{3}$ which gives $\sum{(x^{2}.(1-x)}$ $\leq$ $\frac{6}{27}$ with equality iff x=y=z,

Finally by arithmetic-geometric inequality xyz $\leq$ $(x+y+z)^{3}/27$ = $\frac{1}{27}$ with equality iff x=y=z.

Putting the two inequalities together gives $ \frac{\sum(x.x.(1-x)}_{3}$ + $xyz$ $\leq$ $\frac{7}{27}$ with equality iff x=y=z.
Thus the greatest value of $xy+yz+zx-2xyz$ is $\frac{7}{27}$ and it follows that the least value of $9-18(xy-yz-zx-2xyz)$ is $\frac{13}{3}$

This problem is essentially Q1 from IMO 1984.

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204 posts

lasha

#5 h Sep 1, 2007, 10:52 AM • 2 Y

Y by Adventure10, Mango247

From substituting a=x+y,b=y+z and c=z+x, we get, that x+y+z=3/2. After changing $a^{2}+b^{2}+c^{2}+4/3abc$ by x,y and z, we get:
$a^{2}+b^{2}+c^{2}+4abc/3 = 1/6(27-8xyz)\geq 1/6(27-1) = 13/3$ ,(By Cauchy) with equality if x=y=z=3/2,or a=b=c=1. So min(a^{2}+b^{2}+c^{2}+4abc/3)=13/3.

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566 posts

galav

#6 h Jan 24, 2016, 7:10 PM • 2 Y

Y by Adventure10, Mango247

I think it is true for all $a,b,c$ such that $a+b+c=3$ . They need not be sides of a triangle. My solution ----
Let $f(a,b,c)=a^2+b^2+c^2+\frac{4abc}{3}$
Lemma 1-- At least two of $a,b,c$ are less than $\frac {3}{2}$
Proof-- Suppose not . We get contradiction.
Lemma 2-- Suppose $a$ is less than $\frac {3}{2}$
then $f(a,b,c) \geq f(a,\frac {b+c}{2},\frac {b+c}{2})$
Proof-- Equivalent to $(\frac{1}{2}-\frac{a}{3})(b-c)^2 \geq 0$
Now using Lemma 1 we get $\frac{b+c}{2}$ and $a$ less than $\frac {3}{2}$
Thus we have minimum at $a=b=c$ Thus we get the answer $\frac{13}{3}$

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829 posts

#7 h Jan 24, 2016, 7:15 PM • 2 Y

Y by Adventure10, Mango247

zaya_yc wrote:

5-\frac{6}{9}=\frac{13}{4}$
a=b=c=1

Looks like no one noticed the typo!

This post has been edited 1 time. Last edited by YESMAths, Jan 24, 2016, 7:16 PM

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566 posts

galav

#8 h Jan 24, 2016, 7:18 PM • 2 Y

Y by Adventure10, Mango247

LOL I think they added triangle sides to keep up the computational structure of the exam

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41020 posts

sqing

#9 h Jan 31, 2020, 8:59 AM • 2 Y

Y by AlastorMoody, Adventure10

April wrote:

Let $a,\, b,\, c$ be side lengths of a triangle and $a+b+c = 3$ . Find the minimum of
$a^{2}+b^{2}+c^{2}+\frac{4abc}{3}$

Let $a,b,c$ be positive numbers such that $a+b+c=3.$ Prove that $a^{2}+b^{2}+c^{2}+\frac{4}{3}abc\ge \frac{13}{3} .$ Let $a,b,c$ be positive numbers such that $a+b+c=1.$ Prove that $a^2+b^2+c^2+4abc \ge \frac{13}{27} .$ Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1$ . Prove that $a^2 +b^2 +c^2 +4abc <\frac12$ 1990 ITAMO
Let $a,b,c$ be side lengths of a triangle with $a+b+c=2.$ Prove that $\frac{52}{27}\le a^{2}+b^{2}+c^{2}+2abc<2$ Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1.$ Prove that
${{1}\over{a^2}}+{{1}\over{b^2}}+{{1}\over{c^2}}+{{1}\over{4abc}}\geq \ 32$ S
China North MO 2006:
Let $a,b,c$ are positive numbers such that $a+b+c=3$ , show that: $\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2c^{2}+(a+b)^{2}}\leq 5$

This post has been edited 6 times. Last edited by sqing, Jan 6, 2022, 1:22 AM

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