Let $x,y$ and $z$ be real numbers such that $0 \leq x,y,z \leq 1.$ Prove that \[xyz+(1-x)(1-y)(1-z) \leq 1.\] When does equality hold?
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Tags: probability, function, inequalities, inequalities proposed
14.11.2010 16:02
amparvardi wrote: Let $x,y$ and $z$ be real numbers such that $0 \leq x,y,z \leq 1.$ Prove that \[xyz+(1-x)(1-y)(1-z) \leq 1.\] When does equality hold? $xyz+(1-x)(1-y)(1-z)=1-(x+y+z)+xy+yz+zx\leq 1-(x(1-y)+y(1-z)+z(1-x))\leq1$ $x=y=z=0$ or $x=y=z=1$
14.11.2010 16:06
Alteratively, see that for $x\in\left[0, 1\right]$, we have $x\leqslant\sqrt[3]{x}$, therefore \[\left.xyz+(1-x)(1-y)(1-z)\leqslant \sqrt[3]{xyz}+\sqrt[3]{(1-x)(1-y)(1-z)}\leqslant \frac{x+y+z}{3}+\frac{(1-x)+(1-y)+(1-z)}{3}= 1\mbox{,} \right.\] where the last step is a consequence of AM-GM-inequality.
14.11.2010 16:16
Its equivalent to: $xy+xz+yz \leq x+y+z $ since $xy \leq y$... equality occurs when $x=y=z$
14.11.2010 16:28
amparvardi wrote: Let $x,y$ and $z$ be real numbers such that $0 \leq x,y,z \leq 1.$ Prove that \[xyz+(1-x)(1-y)(1-z) \leq 1.\] When does equality hold? in fact, we have $xyz+(1-x)(1-y)(1-z)+\sum_{cyc}{xy(1-z)}+\sum_{cyc}{(1-x)(1-y)z}=1$ so ...
14.11.2010 16:37
There is no reason why we cannot generalise through Martin N's method (indeed see here ). If $a_1,a_2\ldots a_n$ belong to $(0,1)$ then $a_1a_2\ldots a_n\le \sqrt[n]{a_1a_2\ldots a_n}\le \frac{a_1+a_2+\ldots +a_n}{n}$ and $(1-a_1)(1-a_2)\ldots (1-a_n)\le \sqrt[n]{(1-a_1)(1-a_2)\ldots (1-a_n)}$ $\le \frac{(1-a_1)+(1-a_2)+\ldots +(1-a_n)}{n}$ So adding up we obtain $a_1a_2\ldots a_n+(1-a_1)(1-a_2)\ldots (1-a_n)\le 1$
16.11.2010 21:54
An alternative argument (which underlies kuing's argument): Consider independent events $X_1, \dots X_n$, such that $P(X_i)=a_i$. The probability either all or none of the $X_i$ occur is $a_1a_2 \dots a_n+(1-a_1)(1-a_2)\dots(1-a_n)$, which must be at most $1$. Equality occurs when its impossible to have a nonempty proper subset of the events occurring, which is when either all $a_i$ are $1$ (all events occur) or all $a_i$ are $0$ (no events occur).
17.11.2010 10:20
Am I right about this,please tell me???: LHS is a linear function in each of its variables (that is to say,if I stabilise all the variables except for one,then LHS is a linear function of that variable etc.) Therefore,it assumes its maximum on the endpoints of the intervals in which every variable is defined. Now,we have either that that each variable is 0 or 1.By examining the 8 cases,it is easy to see that the maximum is 1. Ok really,somebody PLEASE tell me if this is correct.I wanna know Regards, Nick
17.11.2010 11:31
We have \[ a_{1}a_{2}\ldots a_{n} \le a_1 \] \[ (1-a_1)(1-a_2)\ldots(1-a_n) \le 1-a_1 \] Adding them up and we yield the desired result.
17.11.2010 12:12
nickthegreek wrote: Am I right about this,please tell me???: LHS is a linear function in each of its variables (that is to say,if I stabilise all the variables except for one,then LHS is a linear function of that variable etc.) Therefore,it assumes its maximum on the endpoints of the intervals in which every variable is defined. Now,we have either that that each variable is 0 or 1.By examining the 8 cases,it is easy to see that the maximum is 1. Ok really,somebody PLEASE tell me if this is correct.I wanna know Regards, Nick You are true about defining defining linear fuction in each variable, but then you need to check just 4 cases: $(0,0,0),(0,0,1),(0,1,1),(1,1,1)$ since it is symmetric.
17.11.2010 12:40
MathUniverse wrote: nickthegreek wrote: Am I right about this,please tell me???: LHS is a linear function in each of its variables (that is to say,if I stabilise all the variables except for one,then LHS is a linear function of that variable etc.) Therefore,it assumes its maximum on the endpoints of the intervals in which every variable is defined. Now,we have either that that each variable is 0 or 1.By examining the 8 cases,it is easy to see that the maximum is 1. Ok really,somebody PLEASE tell me if this is correct.I wanna know Regards, Nick You are true about defining defining linear fuction in each variable, but then you need to check just 4 cases: $(0,0,0),(0,0,1),(0,1,1),(1,1,1)$ since it is symmetric. Thanks!!!
21.11.2010 14:44
equality hold if and only if (x,y,z)=(0,0,0)or (1,1,1).
21.11.2010 15:17
amparvardi wrote: Let $x,y$ and $z$ be real numbers such that $0 \leq x,y,z \leq 1.$ Prove that \[xyz+(1-x)(1-y)(1-z) \leq 1.\] When does equality hold? Let $F(x,y,z)=xyz +(1-x)(1-y)(1-z)$ differentiating with respect to $x$ we get $F' (x,y,z)=yz +(y-1)(1-z)=yz +y -yz -1 +z=y+z-1$ and $F''(x)=0$, thus $F$ is a convex function, and since $F$ is symmetric for all $x,y,z$ then $F$ gets extreme values in boundaries. Therefore $F_{\max} =\{ F(0,0,0),F(1,0,0),F(1,1,0),F(1,1,1) \}=1$, hence $F(x,y,z) \leqslant 1$
01.10.2011 09:15
since the ineq is linear for any of $x,y,z$,so it's monotonous in respect of $x$(or $y,z$).so equality holds only when one of the eight following situations occurs: $(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)$.
03.07.2014 03:28
The following inequality is also true. Let $x_1,x_2,\cdots,x_n$ be real numbers such that $0 \leq x_1,x_2,\cdots,x_n\leq 1.$ Prove that\[x_1x_2\cdots x_n+(1-x^n_1)(1-{x^n_2})\cdots(1-{x^n_n}) \leq 1.\]
03.07.2014 09:06
WLOG $x_1=\max \{x_1,x_2,\cdots,x_n\}$. Then $\prod_{i=1}^{n}x_i\le x_1^n$ and $\prod_{i=1}^{n}(1-x_i)^n\le (1-x_1^n)$. Adding them up finishes the proof.
03.07.2014 09:21
sqing wrote: The following inequality is also true. Let $x_1,x_2,\cdots,x_n$ be real numbers such that $0 \leq x_1,x_2,\cdots,x_n\leq 1.$ Prove that\[x_1x_2\cdots x_n+(1-x^n_1)(1-{x^n_2})\cdots(1-{x^n_n}) \leq 1.\] $x_1x_2\ldots x_n \le \frac{x_1^n + \ldots + x_n^n}{n}$ and $(1 - x_1^n)\ldots(1 - x_n^n) \le \sqrt[n]{(1 - x_1^n)\ldots(1 - x_n^n)} \le \frac{n - x_1^n + \ldots - x_n^n}{n}$. Add.
03.07.2014 10:17
joybangla wrote: WLOG $x_1=\max \{x_1,x_2,\cdots,x_n\}$. Then $\prod_{i=1}^{n}x_i\le x_1^n$ and $\prod_{i=1}^{n}(1-x_i)^n\le (1-x_1^n)$. Adding them up finishes the proof. How is that $\prod_{i=1}^{n}(1-x_i)^n\le (1-x_1^n)$? If we take $x_i = 1/2$ for all $i$ then $\prod_{i=1}^{n}(1-x_i)^n = (1/2)^{n^2}$ and $(1-x_1^n) = 1 - (1/2)^n$ and you can see it's wrong ...
03.07.2014 10:49
mavropnevma wrote: joybangla wrote: WLOG $x_1=\max \{x_1,x_2,\cdots,x_n\}$. Then $\prod_{i=1}^{n}x_i\le x_1^n$ and $\prod_{i=1}^{n}(1-x_i)^n\le (1-x_1^n)$. Adding them up finishes the proof. How is that $\prod_{i=1}^{n}(1-x_i)^n\le (1-x_1^n)$? If we take $x_i = 1/2$ for all $i$ then $\prod_{i=1}^{n}(1-x_i)^n = (1/2)^{n^2}$ and $(1-x_1^n) = 1 - (1/2)^n$ and you can see it's wrong ... Sorry, mavropnevma I made a typo which is quite clear from the problem statement. See that should be written $\prod_{i=1}^{n}(1-x_i^n)\le (1-x_1^n)$. which is true since $x_i\ge 0\implies x_i^n\ge 0\implies (1-x_i^n)\le 1$ now multiplying subsequent inequalities we get what we want. I suppose I am not making any mistakes am I? Do tell me.
03.07.2014 16:04
WakeUp wrote: There is no reason why we cannot generalise through Martin N's method (indeed see here ). If $a_1,a_2\ldots a_n$ belong to $(0,1)$ then$a_1a_2\ldots a_n\le \sqrt[n]{a_1a_2\ldots a_n}\le \frac{a_1+a_2+\ldots +a_n}{n}$ and $(1-a_1)(1-a_2)\ldots (1-a_n)\le \sqrt[n]{(1-a_1)(1-a_2)\ldots (1-a_n)}$ $\le \frac{(1-a_1)+(1-a_2)+\ldots +(1-a_n)}{n}$ So adding up we obtain $a_1a_2\ldots a_n+(1-a_1)(1-a_2)\ldots (1-a_n)\le 1$ If $a_1,a_2\ldots a_n \in [0,1]$ then $a_1a_2\ldots a_n+(1-a_1)(1-a_2)\ldots (1-a_n)\le a_1a_2+(1-a_1)(1-a_2)$ $\le \sqrt{a_1a_2}+\sqrt{(1-a_1)(1-a_2)}\le 1.$
03.07.2014 19:01
sqing wrote: The following inequality is also true. Let $x_1,x_2,\cdots,x_n$ be real numbers such that $0 \leq x_1,x_2,\cdots,x_n\leq 1.$ Prove that\[x_1x_2\cdots x_n+(1-x^n_1)(1-{x^n_2})\cdots(1-{x^n_n}) \leq 1.\] Generalization Let ${{x}_{1}},{{x}_{2}},\cdots ,{{x}_{n}},{{y}_{1}},{{y}_{2}},\cdots ,{{y}_{n}}$ be real numbers such that $0\le {{x}_{1}},{{x}_{2}},\cdots ,{{x}_{n}},{{y}_{1}},{{y}_{2}},\cdots ,{{y}_{n}}\le 1$ and ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}\le \frac{{{y}_{1}}+{{y}_{2}}+\cdots +{{y}_{n}}}{n}.$Prove that ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+(1-{{y}_{1}})(1-{{y}_{2}})\cdots (1-{{y}_{n}})\le 1.$ Proof: ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+(1-{{y}_{1}})(1-{{y}_{2}})\cdots (1-{{y}_{n}})\le {{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+\sqrt[n]{(1-{{y}_{1}})(1-{{y}_{2}})\cdots (1-{{y}_{n}})}\le $ ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+\frac{(1-{{y}_{1}})+(1-{{y}_{2}})+\cdots +(1-{{y}_{n}})}{n}$ $=1+{{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}-\frac{{{y}_{1}}+{{y}_{2}}+\cdots +{{y}_{n}}}{n}\le 1$