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 $x=y=z=1$

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.

Its equivalent to: $xy+xz+yz \leq x+y+z$ since $xy \leq y$... equality occurs when $x=y=z$

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 ...

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 $a_1a_2\ldots a_n+(1-a_1)(1-a_2)\ldots (1-a_n)\le 1$

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).

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

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.

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.

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!!!

equality hold if and only if （x，y，z）＝(0,0,0)or (1,1,1).

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 $F(x,y,z) \leqslant 1$

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: $(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)$.

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 $$x_1x_2\cdots x_n+(1-x^n_1)(1-{x^n_2})\cdots(1-{x^n_n}) \leq 1.$$

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.

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.

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 ...

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.

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 $a_1a_2\ldots a_n+(1-a_1)(1-a_2)\ldots (1-a_n)\le a_1a_2+(1-a_1)(1-a_2)$ $\le \sqrt{a_1a_2}+\sqrt{(1-a_1)(1-a_2)}\le 1.$

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: ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+(1-{{y}_{1}})(1-{{y}_{2}})\cdots (1-{{y}_{n}})\le {{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+\sqrt[n]{(1-{{y}_{1}})(1-{{y}_{2}})\cdots (1-{{y}_{n}})}\le$ ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}+\frac{(1-{{y}_{1}})+(1-{{y}_{2}})+\cdots +(1-{{y}_{n}})}{n}$ $=1+{{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}-\frac{{{y}_{1}}+{{y}_{2}}+\cdots +{{y}_{n}}}{n}\le 1$