$a+b+c=3$

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Source: 5-th Hong Kong Mathematical Olympiad 2002

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3595 posts

#1 h Jan 10, 2007, 5:33 PM • 2 Y

Y by Adventure10, Mango247

Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$ . Prove that $ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$ and find cases of equality.

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#2 h Jan 10, 2007, 6:20 PM • 1 Y

Y by Adventure10

i remember similar problem (but definately not the same and it's likely that solutions is completely different). the difference was that $a+b+c=1$ and $LHS \leq 4/27$

This post has been edited 1 time. Last edited by gollywog, Jan 11, 2007, 12:32 AM

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#3 h Jan 10, 2007, 6:27 PM • 1 Y

Y by Adventure10

http://www.kalva.demon.co.uk/canada/casoln/csol995.html here?

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#4 h Jan 8, 2012, 7:21 PM • 2 Y

Y by Adventure10, Mango247

// there was attempt to prove that $a^2b + b^2c + c^2a \le \frac{27}{8}$ but my proof was wrong

I wonder if it's true

This post has been edited 4 times. Last edited by timon92, Jan 8, 2012, 7:36 PM

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#5 h Jan 8, 2012, 7:26 PM • 3 Y

Y by AdBondEvent, Adventure10, Mango247

You would also be able to use Lagrange Multipliers, since we have the additional condition that $a+b+c = 3$ .

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#6 h Jan 8, 2012, 8:03 PM • 2 Y

Y by Adventure10, Mango247

N.T.TUAN wrote:

Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$ . Prove that $ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$ and find cases of equality.

After homogenization, it's equivalent to:

$\sum a^3+3\sum a^2b+6abc \ge 5\sum ab^2$

$\Leftrightarrow (a+b-c)(a-b)^2 + (4a-4b+c)(a-c)(b-c) + 3abc \ge 0$

What is obviously true by condition. Equality holds when $a=b=\frac{3}{2}$ and $c=0$ . $\blacksquare$

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#7 h Jan 9, 2012, 5:05 AM • 1 Y

Y by Adventure10

gollywog wrote:

i remember similar problem (but definately not the same and it's likely that solutions is completely different). the difference was that $a+b+c=1$ and $LHS \leq 4/27$

Solution. Let $f(x;y;z)=x^2y+y^2z+z^2x.$ WE wish to determine where $f$ is maximal. Since $f$ is cyclic Wlog we may assume that $x=max(x;y;z).$ Since $f(x;y;z)-f(x;z;y)=x^2y+y^2z+z^2x-x^2z-z^2y-y^2x=(y-z)(x-z)(x-y)$ we may also assume $y\geq z.$
Then $f(x+z;y;0)-f(x;y;z)=(x+z)^2y-x^2y-y^2z-z^2x=z^2x+yz(x-y)+xz(y-z)\geq 0$
So we may now assume z=0 The rest follows from AM-GM inequality
$f(x;y;0)=\frac{2x^2y}{2}\leq \frac{1}{2}({\frac{x+x+2y}{3})^3=\frac{4}{27}.}$
Equality occurs when $x=2y$ , hence when $(x;y;z)$ equals $(\frac{2}{3};\frac{1}{3};0), (\frac{1}{3};0;\frac{2}{3})$ or $(0;\frac{2}{3};\frac{1}{3})$ .

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#8 h Jun 2, 2015, 3:23 AM • 2 Y

Y by Adventure10, Mango247

I think the inequality is quite weak.
Well I substituted (a+b+c)^3 = 27 and then applied the Schurhead inequality to get the answer.

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#9 h Jun 2, 2015, 4:51 AM • 2 Y

Y by Adventure10, Mango247

For #1
With conditions $a\ge{b}\ge{c}\ge{0}$ and $a+b+c=3$ ,we have
$\frac{27}{8}-(ab^2+bc^2+ca^2)=\frac{1}{2}(a-b)(a-c)(b-c)+\frac{1}{8}(a+b-c)(a-b)^2+\frac{1}{8}c(a-c)(b-c)+\frac{3}{8}abc\ge{0}$

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#10 h Jun 2, 2015, 11:47 AM • 2 Y

Y by Adventure10, Mango247

vedantbonde wrote:

I think the inequality is quite weak.
Well I substituted (a+b+c)^3 = 27 and then applied the Schurhead inequality to get the answer.

And I always thought Schur/Muirhead only work for symmetric inequalities while this one is cyclic...
Looking forward to see your full solution!

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Quantum_fluctuations

1282 posts

Quantum_fluctuations

#11 h Jul 25, 2020, 9:26 AM • 1 Y

Y by Newmaths

N.T.TUAN wrote:

Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$ . Prove that $ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$ and find cases of equality.

The given inequality is equivalent to,

$\frac{27}{8} \geq ab^2+bc^2+ca^2 \Leftrightarrow (a+b+c)^3 \geq 8(ab^2+bc^2+ca^2)$
$\Leftrightarrow a^3+b^3+c^3+6abc + 3(ab^2+bc^2+ca^2) +3(a^2b+b^2c+c^2a) \geq 8(ab^2+bc^2+ca^2)$
Using schur's inequality, we have

$a^3+b^3+c^3+3abc \geq (ab^2+bc^2+ca^2)+(a^2b+b^2c+c^2a)$
Therefore, it suffices to show that

$(ab^2+bc^2+ca^2)+(a^2b+b^2c+c^2a) + 3(ab^2+bc^2+ca^2) +3(a^2b+b^2c+c^2a) \geq 8(ab^2+bc^2+ca^2)$
$\Leftrightarrow 4(a^2b+b^2c+c^2a) - 4(ab^2+bc^2+ca^2) \geq 0$
$\Leftrightarrow 4(a-b)(b-c)(c-a) \geq 0$
The last inequality is true since we are given $a \geq b \geq c$

Hence, the desired inequality is also true.

Equality occues iff $(a,b,c) =(1,0,0), (0,1,0) , (0,0,1)$ .

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sqing

#12 h Jul 25, 2020, 11:12 AM

Y by

gollywog wrote:

i remember similar problem (but definately not the same and it's likely that solutions is completely different). the difference was that $a+b+c=1$ and $LHS \leq 4/27$

https://artofproblemsolving.com/community/c6h497887p2797425

N.T.TUAN wrote:

Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$ . Prove that $ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$ and find cases of equality.

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