Let $ a,b,c,d \in [0,1]. $ Prove that $$a+b+c+d-ab-bc-cd-da\leq 2$$$$a+b+c+d-abc-bcd-cda-dab\leq 2$$

Let $ a,b,c,d \in [0,1]. $ Prove that $$a^2+b^2+c^2+d^2-ab-bc-cd-da \leq a+b+c+d-ab-bc-cd-da\leq 2$$

Let $ a_1,a_2, \cdots ,a_n \in [0,1] $ $(n\geq 2).$ Prove that $$a^2_1+a^2_2+\cdots+a^2_{n-1}+a^2_n-a_1a_2 -a_2a_3 -\cdots-a_{n-1}a_n -a_na_1 $$$$ \leq a_1+a_2+\cdots+a_{n-1}+a_n-a_1a_2 -a_2a_3 -\cdots-a_{n-1}a_n -a_na_1\leq \frac{n}{2}$$

sqing wrote: Let $ a_1,a_2, \cdots ,a_n \in [0,1] $ $(n\geq 2).$ Prove that $$a^2_1+a^2_2+\cdots+a^2_{n-1}+a^2_n-a_1a_2 -a_2a_3 -\cdots-a_{n-1}a_n -a_na_1 $$$$ \leq a_1+a_2+\cdots+a_{n-1}+a_n-a_1a_2 -a_2a_3 -\cdots-a_{n-1}a_n -a_na_1\leq \frac{n}{2}$$ Solution. Put $a_{n+1}=a_1$. Applying the assumption $a_i\in[0,1]\,\left(i=1,2,\cdots,n\right)$ we get \begin{align*} \sum_{i=1}^na_i-\sum_{i=1}^na_ia_{i+1}=&\sum_{i=1}^na_i(1-a_{i+1})\\ \left(\because a_i\in[0,1]\right)\,\,\le& \sum_{i=1}^n\sqrt{a_i(1-a_{i+1})}\\ \le&\sum_{i=1}^n\frac{1+a_i-a_{i+1}}{2}=\frac{n}{2}. \,\, \blacksquare \end{align*}Edit: Thank @Marrelia for pointing out the mistake in my proof (see #7). I have deleted the wrong conclusion.

ytChen wrote: sqing wrote: Let $ a_1,a_2, \cdots ,a_n \in [0,1] $ $(n\geq 2).$ Prove that $$a^2_1+a^2_2+\cdots+a^2_{n-1}+a^2_n-a_1a_2 -a_2a_3 -\cdots-a_{n-1}a_n -a_na_1 $$$$ \leq a_1+a_2+\cdots+a_{n-1}+a_n-a_1a_2 -a_2a_3 -\cdots-a_{n-1}a_n -a_na_1\leq \frac{n}{2}$$ More generally, $\forall a_i\in[0,1]$, $i=1,2,\cdots,n\,\left(n\ge3\right)$, $$\sum_{i=1}^na_i-\sum_{i=1}^na_ia_{i+1}\le \sum_{i=1}^na_i-\sum_{i=1}^na_ia_{i+1}a_{i+2}\le\frac{n}{2},$$where $a_{n+1}=a_1$ and $a_{n+2}=a_2$. Solution. We first use the assumption $a_i\in[0,1]\,\left(i=1,2,\cdots,n\right)$ to show that $\sum_{i=1}^na_i-\sum_{i=1}^na_ia_{i+1}\le\frac{n}{2}$. Indeed, \begin{align*} \sum_{i=1}^na_i-\sum_{i=1}^na_ia_{i+1}=&\sum_{i=1}^na_i(1-a_{i+1})\\ \left(\because a_i\in[0,1]\right)\,\,\le& \sum_{i=1}^n\sqrt{a_i(1-a_{i+1})}\\ \le&\sum_{i=1}^n\frac{1+a_i-a_{i+1}}{2}=\frac{n}{2}. \end{align*}Using the result just proven above, we get \begin{align*} \sum_{i=1}^na_i-\sum_{i=1}^na_ia_{i+1} a_{i+2} =&\sum_{i=1}^na_i(1-a_{i+1} a_{i+2})\\ \left(\because a_i\in[0,1]\right)\quad\le& \sum_{i=1}^n\sqrt[3]{a_i(1-a_{i+1} a_{i+2})}\\ \le&\sum_{i=1}^n\frac{1+a_i-a_{i+1}a_{i+2}}{3}\\ \left(\because a_{n+1}=a_1\wedge a_{n+2}=a_2\right)\,=&\frac{n}{3}+ \frac{1}{3} \left(\sum_{i=1}^na_i-\sum_{i=1}^n a_ia_{i+1}\right)\\ \le& \frac{n}{3}+ \frac{1}{3}\cdot\frac{n}{2}= \frac{n}{2}.\,\, \blacksquare \end{align*} How do you get ${\sum_{i=1}^n\sqrt[3]{a_i(1-a_{i+1} a_{i+2})}\\ \le\sum_{i=1}^n\frac{1+a_i-a_{i+1}a_{i+2}}{3}}$ I think right hand side should be ${\sum_{i=1}^n\frac{2+a_i-a_{i+1}a_{i+2}}{3}}$

sqing wrote: Let $ a,b,c,d \in [0,1]. $ Prove that $$a+b+c+d-ab-bc-cd-da\leq 2$$$$a+b+c+d-abc-bcd-cda-dab\leq 2$$

Let $ a,b,c,d \in [0,1]. $ Prove that $$a+b+c+d-ab-cd \le 2$$

Let $ a,b,c,d \in [0,1]. $ Prove that $$a^2+b^2+c^2+d^2-ab-cd \leq 2$$

MS_Kekas wrote: For real numbers $a, b, c, d \in [0, 1]$, find the largest possible value of the following expression: $$a^2+b^2+c^2+d^2-ab-bc-cd-da$$ Proposed by Mykhailo Shtandenko this is $\frac{(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2}{2}$. the maximum of this is clearly $2$ as $|a-b|\le 1$ with equality at $(1,0,1,0)$

Let $ a,b,c,d \in [0,1]. $ Prove that $$a^2+b^2+c^2+d^2-\frac{1}{2}(ab+bc+cd+da)\leq 2$$$$a+b+c+d-\frac{1}{2}(ab+bc+cd+da)\leq 2$$$$a^2+b^2+c^2+d^2-ab-\frac{1}{2}cd \leq \frac{5}{2}$$

sqing wrote: Let $ a,b,c,d \in [0,1]. $ Prove that $$a^2+b^2+c^2+d^2-\frac{1}{2}(ab+bc+cd+da)\leq 2$$$$a+b+c+d-\frac{1}{2}(ab+bc+cd+da)\leq 2$$$$a^2+b^2+c^2+d^2-ab-\frac{1}{2}cd \leq \frac{5}{2}$$

sqing wrote: Let $ a,b,c,d \in [0,1]. $ Prove that $$a+b+c+d-\frac{1}{2}(ab+bc+cd+da)\leq 2$$ $$(1-b)(a-1) \le 0\implies a+b-ab \le 1$$$$a+b-ab \le 1,b+c-bc \le 1.c+d-cd \le 1,d+a-da \le 1$$$$\implies$$$$ a+b+c+d-\frac{1}{2}(ab+bc+cd+da)\leq 2$$Thank Marrelia.