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Interesting inequality
sqing   1
N 21 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 2.$ Prove that
$$ (a+1)(b+1)(c +1)-\frac{9}{4}abc\leq 9$$$$ (a+2)(b+2)(c +2)-4 abc\leq 32$$$$ (a+2)(b+2)(c +2)-\frac{17}{4}a b c\leq 30$$$$ (a+1)(b+1)(c +1)-\frac{23}{10}abc\leq\frac{43}{5}$$
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sqing
2 hours ago
sqing
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Interesting inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq \frac{1}{3}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8 . $ Prove that
$$ ab+bc +ca\leq 17+2\sqrt{73}$$Let $ a,b,c\geq \frac{1}{2}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8 . $ Prove that
$$ ab+bc +ca\leq \frac{469+115\sqrt{17}}{32}$$Let $ a,b,c\geq \frac{1}{5}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8 . $ Prove that
$$ ab+bc +ca\leq \frac{569+34\sqrt{281}}{25}$$
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sqing
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Problem for VASC, SI Book
hungkhtn   21
N Wednesday at 4:29 PM by imnotgoodatmathsorry
Source: please let him prove it first
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.\]
21 replies
hungkhtn
Jun 5, 2007
imnotgoodatmathsorry
Wednesday at 4:29 PM
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Problem for VASC, SI Book
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inequalities
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Source: please let him prove it first
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hungkhtn
1750 posts
hungkhtn
#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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Vasc
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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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hungkhtn
1750 posts
hungkhtn
#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. :blush:
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Sunjee
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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arqady
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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hungkhtn
1750 posts
hungkhtn
#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 :lol:
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Sunjee
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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manlio
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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. :wink:
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Sunjee
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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. :wink:
First sorry my bad english, Second watch my proof again. :lol:
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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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Sunjee
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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. :blush: 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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Sunjee
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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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manlio
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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 $2a=3-b$ :wink:
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Sunjee
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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 $2a=3-b$ :wink:
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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Ji Chen
827 posts
Ji Chen
#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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bitrak
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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sqing
41135 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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sqing
41135 posts
sqing
#21 h  Mar 16, 2025, 1:49 AM
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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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