Half concave and half convex theorem

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Source: zhaobin

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#1 h Dec 8, 2005, 3:55 AM • 20 Y

Y by mahanmath, jatin, alibez, zabihpourmahdi, fractals, Grotex, daisyxixi, adityaguharoy, Adventure10, Mango247, and 10 other users

Hey guys,here is my new idea.If it is right.it may be easy.but I think it will be nice,useful and great.

I hope somebody can intreast in my idea,thanks

.And I wonder if others think it before

theorem1 $x_1,x_2,\cdots,x_n$ are n real numbers,such that
$(i),x_1 \le x_2 \le \cdots \le x_n$
$(ii), x_1 , x_2 , \cdots , x_n \in \left[ a ,b \right]$
$(iii)x_1+x_2+\cdots+x_n=C$ (C is contant)
and $f$ is a function on $\left[a,b\right]$ ,if $f$ convave on $\left[ a,c \right]$ and convex on $\left[c,b\right]$
LET $F=f(x_1)+f(x_2)+\cdots+f(x_n)$
then: $F$ is minimal for $x_1=x_2=\cdots=x_{k-1}=a,x_{k+1}=\cdots=x_n$ ( $k=1,2\cdots,n$ )
and $F$ is maximal for $x_1=x_2=\cdots=x_{k-1},x_{k+1}=\cdots=x_n=b$ ( $k=1,2\cdots,n$ )
Proof:just for the minimal(maximal is simalar).by induction.
If there is no ${x_1,x_2,\cdots,x_n}$ or just $x_1 \in \left[a,c\right]$ ,then the theorem is certainly true.
this is because $x_2,x_3,\cdots,x_n \in \left[c,b\right]$ so that $f(x_1)+f(x_2)+\cdots+f(x_n) \ge f(x_1)+(n-1)f(\frac{x_2,x_3,\cdots,x_n}{n-1})$
if there have $x_1,x_2,\cdots,x_i \in \left[a,c\right]$
case(i) $x_1+x_2+\cdots+x_i-(i-1)a <c$ ,we can get
$f(x_1)+f(x_2)+\cdots+f(x_i) \ge (i-1)f(a)+f(x_1+x_2+\cdots+x_i-(i-1)a)$
case(ii) assume $m,(1 \le m \le i)$ be the minimum interger such that $x_1+x_2+\cdots+x_m-(m-1)a \ge c$
then we obtian
$f(x_1)+f(x_2)+\cdots+f(x_m) \ge (m-1)f(a)+f(x_1+x_2+\cdots+x_{m-1}-(m-2)a)+f(x_m) \ge (m-1)f(a)+f(x_1+x_2+\cdots+x_{m-1}+x_m-c-(m-2)a)+f(c)$
we reduce it to be $i-1$ case.
so the theorem is proven.
theorem1' $x_1,x_2,\cdots,x_n$ are n real numbers,such that
$(i),x_1 \le x_2 \le \cdots \le x_n$
$(ii), x_1 , x_2 , \cdots , x_n \in \left[a,b\right]$
$(iii)x_1+x_2+\cdots+x_n=C$ (C is contant)
and $f$ is a function on $\left[a,b\right]$ ,if $f$ convex on $\left[a,c\right]$ and concave on $\left[c,b\right]$
LET $F=f(x_1)+f(x_2)+\cdots+f(x_n)$
then: $F$ is minimal for $x_1=x_2=\cdots=x_{k-1},x_{k+1}=\cdots=x_n=b$ ( $k=1,2\cdots,n$ )
and $F$ is maximal for $x_1=x_2=\cdots=x_{k-1}=a,x_{k+1}=\cdots=x_n$ ( $k=1,2\cdots,n$ )
proof is similar to theorem1.
Application
1.http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1%2B%5C%5Ccos&t=59649
Let ABC be an acute-angled triangle,Prove:
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$
proof:just let ${f(x)=\cos^2{x}}{\cos{x}+1}$ you will find $f$ satisfy the condtion with theorem1',
so we just need to prove :
$\frac{\cos^2{A}}{\cos{A}+1}+\frac{\cos^2{B}}{\cos{B}+1}\ge\frac{1}{2}(A+B=\frac{\pi}{2})$
or $\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$ ( $A=B$ )

2..http://www.mathlinks.ro/Forum/viewtopic.php?t=64122
$a,b,c \ge 0$
$\sqrt{1+\frac{48a}{b+c}}+\sqrt{1+\frac{48b}{c+a}}+\sqrt{1+\frac{48c}{a+b}} \ge 15$
proof:let $x=\frac{a}{a+b+c}$ and y,z similar
notice $\sqrt{1+\frac{48a}{b+c}}=\sqrt{\frac{48}{1-x}-47}$
and let $f(t)=\sqrt{\frac{48}{1-t}-47}$ you will find $f$ satisfy the condtion with theorem1,
wlog $x \le y \le z$ then we should only to prove
$f(y)+f(z) \ge 15$ ( $y+z=1$ ) or $f(x)+f(y)+f(z) \ge 15$ ( $y=z$ )
theorem2 $x_1,x_2,\cdots,x_n$ are n real numbers,such that
$(i),x_1 \le x_2 \le \cdots \le x_n$
$(ii), x_1 , x_2 , \cdots , x_n \in \left( -\infty ,+\infty \right)$
$(iii)x_1+x_2+\cdots+x_n=C$ (C is contant)
and $f$ is a function on $\left( -\infty ,+\infty \right)$ ,if $f$ convave on $\left(-\infty,c \right]$ and convex on $\left[c,+\infty \right)$
LET $F=f(x_1)+f(x_2)+\cdots+f(x_n)$
then: $F$ is minimal for $x_{2}=x_3=\cdots =x_n$
and $F$ is maximal for $x_1=x_2=\cdots =x_{n-1}$
Proof:just for the minimal(maximal is simalar).
assume $x_1,x_2,\cdots,x_i \in \left(-\infty,c \right]$
because $f$ convave on $\left(-\infty,c \right]$
then we get $f(x_1)+f(x_2)+\cdots+f(x_i) \ge (i-1)f(c)+f(x_1+x_2+\cdots+x_i-(i-1)c)$
and $(i-1)f(c)+f(x_{i+1})+f(x_{i+2})+\cdots+f(x_{n}) \ge (n-1)f(\frac{(i-1)c+x_{i+1}+\cdots+x_n}{n-1})$
so $f(x_1)+f(x_2)+\cdots+f(x_n) \ge (n-1)f(\frac{(i-1)c+x_{i+1}+\cdots+x_n}{n-1})+f(x_1+x_2+\cdots+x_i-(i-1)c)$
so the theorem2 is proved.
theorem2' $x_1,x_2,\cdots,x_n$ are n real numbers,such that
$(i),x_1 \le x_2 \le \cdots \le x_n$
$(ii), x_1 , x_2 , \cdots , x_n \in \left( -\infty ,+\infty \right)$
$(iii)x_1+x_2+\cdots+x_n=C$ (C is contant)
and $f$ is a function on $\left( -\infty ,+\infty \right)$ ,if $f$ convex on $\left(-\infty,c \right]$ and concave on $\left[c,+\infty \right)$
LET $F=f(x_1)+f(x_2)+\cdots+f(x_n)$
then: $F$ is minimal for $x_{1}=x_3=\cdots =x_{n-1}$
and $F$ is maximal for $x_2=x_2=\cdots =x_{n}$

Application
http://www.mathlinks.ro/Forum/viewtopic.php?t=32031
http://www.mathlinks.ro/Forum/viewtopic.php?t=64793
$x,y,z \ge 0$
find the minimum of $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}$
proof:let $f(t)=\frac 1{(1+e^t)^k}$ then we calculate that
$f\"(t)=\frac{e^x(k(k+1)e^x-k)}{(1+e^x)^{k+2}}$
which impies it can use theorem2 .
Wlog $x \le y \le z$
by theorem2 I think we should only consider the case $y=z$
I think it can also work when find the maximum of $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}$ .

And I think there may be another more applications,If you find,please write it here,thanks

,and If there is something wrong in my post please point it out for me.thanks :)and welcome any advice.
(and I find the theorem2 is a little similar to VASC's Right-Convex Function Theorem )

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Vasc

#2 h Dec 14, 2005, 12:41 PM • 8 Y

Y by shima, bararobertandrei, Adventure10, Mango247, and 4 other users

I agree on Theorem 1 and 1'.
With regard to Theorem 2 and 2', there is a formulation trouble (problem), because it is possible that $F$ is not minimal/maximal.

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#3 h Dec 14, 2005, 2:41 PM • 6 Y

Y by Adventure10, Mango247, and 4 other users

thanks very much,VASC.
can we add a condition that $f(x)$ upbound or lowbound

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Vasc

#4 h Dec 15, 2005, 9:48 AM • 6 Y

Y by Adventure10, Mango247, and 4 other users

I think we can use a more general formulation. For example,
Theorem2. ...
a) To obtain the minimal value or the greatest lower bound of $F$ , it suffices to consider $x_2=x_3=...=x_n$ .

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#5 h Dec 15, 2005, 2:57 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

Vasc wrote:

I think we can use a more general formulation. For example,
Theorem2. ...
a) To obtain the minimal value or the greatest lower bound of $F$ , it suffices to consider $x_2=x_3=...=x_n$ .

yes,you are right.It will be nicer

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Vasc

#6 h Dec 17, 2005, 4:06 PM • 7 Y

Y by Adventure10, Mango247, and 5 other users

I think we can extend theorems 1-1' as follows:

Theorem 1. Let $a,b$ be fixed reals ( $a<b$ ), and let $x_1,x_2,...,x_n$ be real numbers such that
$a\leq x_1\leq x_2\leq... \leq x_n \leq b$ and $x_1+x_2+...+x_n=constant$ .
Let $f$ be a function either continuous on $[a,b]$ , or continuous on $[a,b)$ and with $\lim_{x\rightarrow b}{f(x)}=\infty$ . Moreover,
$f$ is concave on $[a,c]$ and convex on $[c,b)$ , where $c\in(a,b)$ .
Then, the expression $F=f(x_1)+f(x_2)+...+f(x_n)$ is minimal for $x_1=...=x_{k-1}=a$ and $x_{k+1}=...=x_n>c$ ,
where $k\in(1,2,...,n)$ .

Theorem 2. Let $a,b$ be fixed reals ( $a<b$ ), and let $x_1,x_2,...,x_n$ be real numbers such that
$a\leq x_1\leq x_2\leq... \leq x_n \leq b$ and $x_1+x_2+...+x_n=constant$ .
Let $f$ be a function either continuous on $[a,b]$ , or continuous on $(a,b]$ and with $\lim_{x\rightarrow a}{f(x)}=-\infty$ . Moreover,
$f$ is concave on $(a,c]$ and convex on $[c,b]$ , where $c\in(a,b)$ .
Then, the expression $F=f(x_1)+f(x_2)+...+f(x_n)$ is maximal for $x_1=...=x_{k-1}<c$ and $x_{k+1}=...=x_n=b$ ,
where $k\in(1,2,...,n)$ .

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#7 h Dec 18, 2005, 3:37 AM • 6 Y

Y by Adventure10, Mango247, and 4 other users

thanks.Vasc.
but I think the condition with continuous is not necessary.
because f concave on $[a,c]$ and convex on $[c,b)$ contain f continuous on $(a,c)$ and $(c,b)$
and more:if $f$ is not continuous on $a$ or $b$ ,the theorem is still true

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#8 h Dec 18, 2005, 9:01 AM • 5 Y

Y by Adventure10 and 4 other users

now I find two Applications of the theorem 1 or 1'.
$1$ ,find the minamum the $k>0$ such that for every $a,b,c \ge 0$
we have $\sqrt[k]{\frac a {b+c}}+\sqrt[k]{\frac b {c+a}}+\sqrt[k]{\frac c {a+b}} \ge 3(\frac 1 2)^k$
the answer is $\frac{\ln 3}{\ln 2}-1$
and I think it is a famous one,I haven't see a solution without my theorem before

.Now I think I can slove it,but also cost a lot of calclulations

.
$2$ ,find the minamum the $k>0$ such that for every $a,b,c \ge 0,a+b+c=3$
we have $\sqrt[k] a +\sqrt[k] b +\sqrt[k] c \ge ab+bc+ca$
it is a generaltion of russian one.It have posted on the forum.As far as I know,no one give a solution.
also it cost a lot of calclulations

.

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#9 h Dec 18, 2005, 5:54 PM • 4 Y

Y by Adventure10 and 3 other users

zhaobin wrote:

Hey guys,here is my new idea.If it is right.it may be easy.but I think it will be nice,useful and great.

I hope somebody can intreast in my idea,thanks

.And I wonder if others think it before

I already knew these theorems from mathlinks (Mildorf). It also appears in his inequalities pdf.

An alternative proof would be by Karamata without induction.

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#10 h Dec 19, 2005, 1:41 AM • 5 Y

Y by Adventure10, Mango247, and 3 other users

perfect_radio wrote:

zhaobin wrote:

Hey guys,here is my new idea.If it is right.it may be easy.but I think it will be nice,useful and great.

I hope somebody can intreast in my idea,thanks

.And I wonder if others think it before

I already knew these theorems from mathlinks (Mildorf). It also appears in his inequalities pdf.

An alternative proof would be by Karamata without induction.

sorry,if it is ture.

can you upload Mildorf's inequalities pdf?(I can't find it)

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#11 h Dec 19, 2005, 9:12 AM • 5 Y

Y by Adventure10, Mango247, and 3 other users

See here : http://web.mit.edu/~tmildorf/www/Inequalities.pdf . (Problem 21) It's more or less your theorem

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#12 h Dec 19, 2005, 1:46 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

ok.thanks,

and I come out the idea when I sloving the inequality:
$\sqrt{1+\frac{48a}{b+c}}+\sqrt{1+\frac{48b}{c+a}}+\sqrt{1+\frac{48c}{a+b}} \ge 15$
and
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$

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#13 h Feb 4, 2007, 12:38 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

zhaobin wrote:

If it is right, it may be easy.
theorem1 $x_{1},x_{2},\cdots,x_{n}$ are n real numbers,such that
$(i),x_{1}\le x_{2}\le \cdots \le x_{n}$
$(ii), x_{1}, x_{2}, \cdots , x_{n}\in \left[ a ,b \right]$
$(iii)x_{1}+x_{2}+\cdots+x_{n}=C$ (C is contant)
and $f$ is a function on $\left[a,b\right]$ ,if $f$ convave on $\left[ a,c \right]$ and convex on $\left[c,b\right]$
LET $F=f(x_{1})+f(x_{2})+\cdots+f(x_{n})$
then: $F$ is minimal for $x_{1}=x_{2}=\cdots=x_{k-1}=a,x_{k+1}=\cdots=x_{n}$ ( $k=1,2\cdots,n$ )
and $F$ is maximal for $x_{1}=x_{2}=\cdots=x_{k-1},x_{k+1}=\cdots=x_{n}=b$ ( $k=1,2\cdots,n$ )

This is not ture, for example,

Let $x,y,z$ be nonnegative numbers such that $x+y+z=5$ .

$f(x)=5x^{4}-28x^{3}$ , then $f''(x)=12x(5x-14)$ .

$f(x)$ is convave on $\left[0,\frac{14}{5}\right]$ and convex on $\left[\frac{14}{5},5\right]$ .

$f(0)+f(0)+f(5) =-375$ ,

$f(0)+f(\frac{5}{2})+f(\frac{5}{2})=-\frac{3875}{8}=-484.375$ ,

$f(\frac{5}{3})+f(\frac{5}{3})+f(\frac{5}{3}) =-\frac{7375}{27}=-273.148\cdots$ ,

but

$f(x)+f(y)+f(z) \geq f(0)+f(1)+f(4) =-535$ .

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#14 h Feb 4, 2007, 3:57 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

Apparently, the minimum is reached for $a = x_{1}= \ldots = x_{k}\leq x_{k+1}< c \leq x_{k+2}= \ldots = x_{n}$ .

This post has been edited 2 times. Last edited by perfect_radio, Feb 5, 2007, 10:31 AM

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#15 h Feb 5, 2007, 8:54 AM • 4 Y

Y by Adventure10 and 3 other users

Ji Chen wrote:

zhaobin wrote:

If it is right, it may be easy.
theorem1 $x_{1},x_{2},\cdots,x_{n}$ are n real numbers,such that
$(i),x_{1}\le x_{2}\le \cdots \le x_{n}$
$(ii), x_{1}, x_{2}, \cdots , x_{n}\in \left[ a ,b \right]$
$(iii)x_{1}+x_{2}+\cdots+x_{n}=C$ (C is contant)
and $f$ is a function on $\left[a,b\right]$ ,if $f$ convave on $\left[ a,c \right]$ and convex on $\left[c,b\right]$
LET $F=f(x_{1})+f(x_{2})+\cdots+f(x_{n})$
then: $F$ is minimal for $x_{1}=x_{2}=\cdots=x_{k-1}=a,x_{k+1}=\cdots=x_{n}$ ( $k=1,2\cdots,n$ )
and $F$ is maximal for $x_{1}=x_{2}=\cdots=x_{k-1},x_{k+1}=\cdots=x_{n}=b$ ( $k=1,2\cdots,n$ )

This is not ture, for example,

Let $x,y,z$ be nonnegative numbers such that $x+y+z=5$ .

$f(x)=5x^{4}-28x^{3}$ , then $f''(x)=12x(5x-14)$ .

$f(x)$ is convave on $\left[0,\frac{14}{5}\right]$ and convex on $\left[\frac{14}{5},5\right]$ .

$f(0)+f(0)+f(5) =-375$ ,

$f(0)+f(\frac{5}{2})+f(\frac{5}{2})=-\frac{3875}{8}=-484.375$ ,

$f(\frac{5}{3})+f(\frac{5}{3})+f(\frac{5}{3}) =-\frac{7375}{27}=-273.148\cdots$ ,

but

$f(x)+f(y)+f(z) \geq f(0)+f(1)+f(4) =-535$ .

Note that in the theorem I wrote the $x_{k}$ is not fixed...

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#16 h Feb 5, 2007, 9:31 AM • 3 Y

Y by Adventure10 and 2 other users

zhaobin wrote:

Note that in the theorem I wrote the $x_{k}$ is not fixed...

Indeed, you are right. It's useful.

zhaobin wrote:

Application $a,b,c \ge 0$
$\sqrt{1+\frac{48a}{b+c}}+\sqrt{1+\frac{48b}{c+a}}+\sqrt{1+\frac{48c}{a+b}}\ge 15$
proof:let $x=\frac{a}{a+b+c}$ and y,z similar
notice $\sqrt{1+\frac{48a}{b+c}}=\sqrt{\frac{48}{1-x}-47}$
and let $f(t)=\sqrt{\frac{48}{1-t}-47}$ you will find $f$ satisfy the condtion with theorem1,
wlog $x \le y \le z$ then we should only to prove
$f(y)+f(z) \ge 15$ ( $y+z=1$ ) or $f(x)+f(y)+f(z) \ge 15$ ( $y=z$ )

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kuing

#17 h Sep 13, 2009, 12:51 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

very useful theorem.

Ding Qi...

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kuing

#18 h Dec 5, 2009, 5:03 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

zhaobin wrote:

case(ii) assume $m,(1 \le m \le i)$ be the minimum interger such that $x_1 + x_2 + \cdots + x_m - (m - 1)a \ge c$
then we obtian
$f(x_1) + f(x_2) + \cdots + f(x_m) \ge (m - 1)f(a) + f(x_1 + x_2 + \cdots + x_{m - 1} - (m - 2)a) + f(x_m) \ge (m - 1)f(a) + f(x_1 + x_2 + \cdots + x_{m - 1} + x_m - c - (m - 2)a) + f(c)$
we reduce it to be $i - 1$ case.
so the theorem is proven.

I think
$f(x_1) + f(x_2) + \cdots + f(x_m) \ge (m - 2)f(a) + f(x_1 + x_2 + \cdots + x_{m - 1} - (m - 2)a) + f(x_m) \ge (m - 2)f(a) + f(x_1 + x_2 + \cdots + x_{m - 1} + x_m - c - (m - 2)a) + f(c)$

...

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zdyzhj

#19 h Aug 11, 2011, 11:12 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

for the first step ,how can we deduce that $F$ is minimal for ${{x}_{1}}={{x}_{2}}=\cdots {{x}_{k-1}}=a,{{x}_{k+1}}=\cdots {{x}_{n}}$ as well as the following proof. who can give a explanation about the condtion for the equality.

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