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Problem for VASC, SI Book

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three variable inequality

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Source: please let him prove it first

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1750 posts

#1 h Jun 5, 2007, 4:23 AM • 6 Y

Y by Adventure10, Mango247, and 4 other users

Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$ . Prove that
$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.$

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2861 posts

Vasc

#2 h Jun 5, 2007, 6:07 AM • 12 Y

Y by bobthesmartypants, Adventure10, Mango247, and 9 other users

Use
$\sqrt{1+a^{3}}\le 1+\frac{a^{2}}{2}$ ,
$\sqrt{1+b^{3}}\le 1+\frac{b^{2}}{2}$ ,
$\sqrt{1+c^{3}}\le 1+\frac{c^{2}}{2}$ .

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1750 posts

#3 h Jun 5, 2007, 6:38 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

You are really excellent, VASC. When I created and then solved this problem, I used mixing variable with 2 pages solution.

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525 posts

Sunjee

#4 h Jun 5, 2007, 6:39 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Vasc wrote:

Use
$\sqrt{1+a^{3}}\le 1+\frac{a^{2}}{2}$ ,
$\sqrt{1+b^{3}}\le 1+\frac{b^{2}}{2}$ ,
$\sqrt{1+c^{3}}\le 1+\frac{c^{2}}{2}$ .

How to prove: if $a+b+c=3$ then
$ab^{2}+bc^{2}+ca^{2}\leq 4$

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30150 posts

arqady

#5 h Jun 5, 2007, 7:21 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Wow, Vasc! Ingeniously simply!

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1750 posts

#6 h Jun 5, 2007, 7:32 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Yes ! When I can only find out this solution one week after

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525 posts

Sunjee

#7 h Jun 5, 2007, 7:40 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let me finished my proof:
$f(a,b,c)=ab^{2}+bc^{2}+ca^{2}$
At least one of following inequalities is true
$1.~~f(2a,b,0)\geq f(a,b,c)$

$2.~~f(0,2b,c)\geq f(a,b,c)$

$3.~~f(a,0,2c)\geq f(a,b,c)$
Let $(2a,b,0)< f(a,b,c),(0,2b,c)< f(a,b,c),(a,0,2c)< f(a,b,c),$ . then we have
$ab^{2}+bc^{2}+ca^{2}<0$ which is not true. So we can write
$f(2a,b,0)\geq f(a,b,c)$
and
$f(2a,b,0)=f(3-b,b,0)=\frac{1}{2}(3-b)(3-b)(2b)\leq \frac{1}{2}(\frac{3-b+3-b+2b}{3})^{3}=4$
Inequality holds if $(a,b,c)=(2,1,0)$ .

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3251 posts

manlio

#8 h Jun 6, 2007, 5:36 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Dear Sunjee if $a+b+c=3$ how can you suppose that $a+2b=3$ I don't understand this step, sorry.

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525 posts

Sunjee

#9 h Jun 6, 2007, 6:20 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

manlio wrote:

Dear Sunjee if $a+b+c=3$ how can you suppose that $a+2b=3$ I don't understand this step, sorry.

First sorry my bad english, Second watch my proof again.

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Kunihiko_Chikaya

14512 posts

Kunihiko_Chikaya

#10 h Jun 6, 2007, 6:31 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

hungkhtn wrote:

Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$ . Prove that
$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.$

For $0\leq a,\ b, \ c \leq 3$ , we have $a\sqrt{1+b^{3}}\leq \frac{3}{2\sqrt{2}}ab+\frac{1}{2\sqrt{2}}a$ e.t.c.

$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\leq \frac{3}{2\sqrt{2}}(ab+bc+ca)+\frac{1}{2\sqrt{2}}(a+b+c)$

$\leq \frac{1}{2\sqrt{2}}(a+b+c)^{2}+\frac{1}{2\sqrt{2}}(a+b+c)=3\sqrt{2}\leq 5$ .

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525 posts

Sunjee

#11 h Jun 6, 2007, 6:59 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

[/quote]

$a\sqrt{1+b^{3}}\leq \frac{3}{2\sqrt{2}}ab+\frac{1}{2\sqrt{2}}a$ e.t.c.

[/quote]
I don't understand it. Can you post solution.

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Kunihiko_Chikaya

14512 posts

Kunihiko_Chikaya

#12 h Jun 6, 2007, 7:08 AM • 4 Y

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

Sorry, my solution didn't make a sense.

The graph of $y=\sqrt{1+x^{3}}\ (x\geq-1)$ has an inflection point at $x=0$ . I was trapped by the convexity.

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Kunihiko_Chikaya

14512 posts

Kunihiko_Chikaya

#13 h Jun 6, 2007, 7:53 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$a\sqrt{1+b^{3}}=a\sqrt{(1+b)(1-b+b^{2})}\leq \frac{a(2+b^{2})}{2}$ e.t.c.

$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\leq a+b+c+\frac{1}{2}(ab^{2}+bc^{2}+ca^{2})\leq 3+\frac{1}{2}\cdot 4=5$ .

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525 posts

Sunjee

#14 h Jun 6, 2007, 8:10 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

kunny wrote:

$a\sqrt{1+b^{3}}=a\sqrt{(1+b)(1-b+b^{2})}\leq \frac{a(2+b^{2})}{2}$ e.t.c.

$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\leq a+b+c+\frac{1}{2}(ab^{2}+bc^{2}+ca^{2})\leq 3+\frac{1}{2}\cdot 4=5$ .

How do you prove $ab^{2}+bc^{2}+ca^{2}\leq 4$

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3251 posts

manlio

#15 h Jun 6, 2007, 9:02 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Sorry Sunjee, but I don't think you can set $2a=3-b$

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525 posts

Sunjee

#16 h Jun 6, 2007, 9:07 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

manlio wrote:

Sorry Sunjee, but I don't think you can set $2a=3-b$

At least one of following inequalities is true
$1.~~f(2a,b,0)\geq f(a,b,c)$

$2.~~f(0,2b,c)\geq f(a,b,c)$

$3.~~f(a,0,2c)\geq f(a,b,c)$

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Kunihiko_Chikaya

14512 posts

Kunihiko_Chikaya

#17 h Jun 6, 2007, 11:02 AM • 2 Y

Y by Adventure10 and 1 other user

Sunjee wrote:

kunny wrote:

$a\sqrt{1+b^{3}}=a\sqrt{(1+b)(1-b+b^{2})}\leq \frac{a(2+b^{2})}{2}$ e.t.c.

$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\leq a+b+c+\frac{1}{2}(ab^{2}+bc^{2}+ca^{2})\leq 3+\frac{1}{2}\cdot 4=5$ .

How do you prove
$ab^{2}+bc^{2}+ca^{2}\leq 4$

From $a+b+c=3,\ a\geq 0,\ b\geq 0,\ c\geq 0$ ,

$ab^{2}+bc^{2}+ca^{2}=ab^{2}+b(3-a-b)^{2}+a^{2}(3-a-b)$
$=b^{3}+3(a-2)b^{2}-3(2a-3)b-a^{3}+3a^{2}: =f(b)\ (0\leq b\leq 3)$ .

$f'(b)=3(b-1)(b+2a-3)$ , since $0\leq a\leq 3$ ,

Case 1: $3-2a<1\Longleftrightarrow a>1$ , Remark that $f(3-2a)=f(a)$ , we are to classify the cases as follows.

If $1<a\leq 3$ , we have $f_{max}=f(a)=3a^{3}-9a^{2}+9a: =g(a)$ , yielding $g'(a)=9(a-1)^{2}$
$\geq 0\Longrightarrow g_{max}=g(3)=27$ when $a=3,\ b=3,\ c=-3$ No good!

If $a>3$ , we have $f_{max}=f(3-2a)=a(2a-3)^{2}>27$ .

Case 2: $1<3-2a\Longleftrightarrow a<1\Longrightarrow 4-3a<1$ ,

$f_{max}=f(1)=-a^{3}+3a^{2}-3a+4: =h(a)\Longrightarrow h'(a)=-3(a-1)^{2}\leq 0$
yielding $h_{max}=h(0)=4$ when $a=0,\ b=1,\ c=2$ .

Case 3: $3-2a=1\Longleftrightarrow a=1$ , we have $f(b)=b^{3}-3b^{2}+6b+2$ ,
$\Longrightarrow f'(b)=3(b-1)^{2}\geq 0$ , yielding $f_{max}=f(3)=20$ when $a=1,\ b=3,\ c=-1$ No good!

Therefore $ab^{2}+bc^{2}+ca^{2}\leq 4$ .

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827 posts

#18 h Dec 16, 2007, 1:40 AM • 3 Y

Y by newsun, Adventure10, and 1 other user

hungkhtn wrote:

Let $x,y,z$ be non-negative real numbers such that $x + y + z = 3$ . Prove that
$x\sqrt {1 + y^{3}} + y\sqrt {1 + z^{3}} + z\sqrt {1 + x^{3}}\le 5$ .

Vasc wrote:

Use $\sqrt {1 + y^{3}}\le 1 + \frac {y^{2}}{2}$ .

$[2 + y^{2}]^2 - 4(1 + y^3) = y^2(2 - y)^2\geq0$ .

Sunjee wrote:

How to prove: if $x + y + z = 3$ then $xy^{2} + yz^{2} + zx^{2}\leq 4$ .

$4(x + y + z)^3 - 27(xy^2 + yz^2 + zx^2)$

$= (z + x - 2y)^2(y + 4z - 5x) + 9x(x^2 + y^2 + z^2 + 2yz - zx - xy) \geq 0$ ,

which is clearly true for $x = \min\{x,y,z\}$ .

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935 posts

bitrak

#19 h Feb 9, 2015, 3:59 PM • 2 Y

Y by Adventure10, Mango247

Sunjee wrote:

Vasc wrote:

Use
$\sqrt{1+a^{3}}\le 1+\frac{a^{2}}{2}$ ,
$\sqrt{1+b^{3}}\le 1+\frac{b^{2}}{2}$ ,
$\sqrt{1+c^{3}}\le 1+\frac{c^{2}}{2}$ .

How to prove: if $a+b+c=3$ then
$ab^{2}+bc^{2}+ca^{2}\leq 4$

We will prove stronger inequality.

Attachments:

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41137 posts

sqing

#20 h Feb 9, 2015, 10:33 PM • 2 Y

Y by Adventure10, Mango247

Let $a$ , $b$ , $c$ be three non-negative real numbers such that $a+b+c=3$ . Prove that
$3\le a \sqrt{1+b^3}+b\sqrt{1+c^3} +c \sqrt{1+a^3}\le5$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487447&p=2731584#p2731584
Let $a,b,c \in \mathbb{R^{+}}$ such that $a+b+c=3$ .Prove the inequality $a^2b+b^2c+c^2a+abc\le 4.$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=497887

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41137 posts

sqing

#21 h Mar 16, 2025, 1:49 AM

Y by

Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$ . Prove that
$a\sqrt{1+b^{3}}+kb\sqrt{1+c^{3}}+kc\sqrt{1+a^{3}}\le 5k$ Where $k\geq 1.$

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imnotgoodatmathsorry

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imnotgoodatmathsorry

#22 h Wednesday at 4:29 PM

Y by

Sunjee wrote:

Vasc wrote:

Use
$\sqrt{1+a^{3}}\le 1+\frac{a^{2}}{2}$ ,

$\sqrt{1+b^{3}}\le 1+\frac{b^{2}}{2}$ ,

$\sqrt{1+c^{3}}\le 1+\frac{c^{2}}{2}$ .

How to prove: if $a+b+c=3$ then
$ab^{2}+bc^{2}+ca^{2}\leq 4$

WLOG, we have $(b-a)(b-c) \le 0$ so we have:
$b^2 + ca \le ab+bc$ or $ab^2 +bc^2+ca^2 \le a^2 b + abc + bc^2 \le a^2 b + 2abc + bc^2 = b (a+c)^2$
By $AM-GM$ , we have: $b (a+c)^2 \le \frac{1}{2} \frac{(2a+2b+2c)^3}{27} = 4$
The equality occurs when $(a;b;c) = (1;2;0)$ and its permutation.

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