shoki wrote: Find the least value of $k$ such that for all $a,b,c,d \in \mathbb{R}$ we have \[ \begin{array} c \sqrt{(a^2+1)(b^2+1)(c^2+1)} +\sqrt{(b^2+1)(c^2+1)(d^2+1)} +\sqrt{(c^2+1)(d^2+1)(a^2+1)} +\sqrt{(d^2+1)(a^2+1)(b^2+1)} \\ \ \\ \ge 2( ab+bc+cd+da+ac+bd)-k \end{array}\] From Cauchy Schwarz inequality $\sum \sqrt{(a^2+1)(b^2+1)(c^2+1)}$ $=\sum \sqrt{\big[a^2+1\big]\big[(b+c)^2+(bc-1)^2\big]}$ $\ge \sum \big[a(b+c)+bc-1\big]=2\sum_{sym} ab -4$ if we take $a=b=c=d=\sqrt{3}$ then $k=4$ works..

Very nice! Thanks Mr. Jafari Here is my solution to problem 85: Find minimum real number $k$ such that for all real numbers $a,b,c,d$: $\sum\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}+k\ge2(ab+bc+cd+da+ac+bd)$. (Proposed by Dr.Amir Jafari and Mohammad Jafari) The answer is $4$. First, we assume that the real number $k$ satisfy the inequality: $\sum\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}+k\ge2(ab+bc+cd+da+ac+bd)$ for all real numbers $a,b,c,d$. When $a=b=c=d=\sqrt{3}$ the inequality become $k\ge4$. By Cauchy Schwarz inequality, we have $\sum\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}+4$ $=\sum\sqrt{\left(a^2+1\right)\left((b+c)^2+(bc-1)^2\right)}+4$ $\ge\sum\left(a(b+c)+(1)(bc-1)\right)+4$ $=2(ab+bc+cd+da+ac+bd)$. Hence the minimum of $k$ is $4$.

Let $a=b=c=d$ and $f(a) = 4 \sqrt{(a^2 +1)^3}- 12a^2$ Our goal is find min $f$ , we have $f' {(a)} = 12a ( \sqrt{a^2 +1} -2)$ so we will find $k$ for $a=0$ or $|a|= \sqrt {3}$ It gives us $k \geq 4$ . Now we want to prove the inequality for $k=4$ . In this step it`s natural to prove this symetric inequality : $\sqrt{(a^{2}+1)(b^{2}+1)(c^{2}+1)} +1 \geq ab+ bc +ca$ Which is after squaring boils down to : $(abc)^2 + (a+b+c)^2 \geq 2abc(a+b+c)$ Which is AM-GM .

mahanmath wrote: Let $a=b=c=d$ and $f(a) = 4 \sqrt{(a^2 +1)^3}- 12a^2$ Our goal is find min $f$ , we have $f' {(a)} = 12a ( \sqrt{a^2 +1} -2)$ so we will find $k$ for $a=0$ or $|a|= \sqrt {3}$ It gives us $k \geq 4$ . Now we want to prove the inequality for $k=4$ . In this step it`s natural to prove this symetric inequality : $\sqrt{(a^{2}+1)(b^{2}+1)(c^{2}+1)} +1 \geq ab+ bc +ca$ Which is after squaring boils down to : $(abc)^2 + (a+b+c)^2 \geq 2abc(a+b+c)$ Which is AM-GM . This is also my proof.

$\sqrt{(a^2+1)(b^2+1)(c^2+1)}=\sqrt{(a+b+c-abc)^2+(ab+bc+ac-1)^2} \geq ab+bc+ac-1$ Now summation over cyclic permutation gives the result for $k=4$ and equality for $a+b+c=abc$.

First note that there are no restriction on $a,b,c,d$ except being as $+ve$ reals, now suppose for minimum $f(a,b,c)$ we've $\sqrt{(a^2+1)(b^2+1)(c^2+1)}+f(a,b,c)\geq (ab+bc+ac)$. Certainly as there are no restriction on $a,b,c,d$ so we must have $f(a,b,c)=f(b,c,d)=f(c,d,a)=f(d,a,b)$. Now suppose $f(c)=\sqrt{ (a^2+1)(b^2+1)(c^2+1)}+f(a,b,c)-(ab+bc+ac)$ then if $abc\geq a+b+c$ we get $f$ is increasing, so $f(c)\geq f(0)=\sqrt{(a^2+1)(b^2+1)}-ab\geq +f(a,b,c)=(a-b)^2+f(a,b,c)-1$. So $a=b\implies f(a,b,c)=1$. Now for other case we get $f$ is decreasing , then $f(c)>f(\frac {a+b}{ab-1})$,now since now we've equation in two variables, just by easy calculus again we get $f(a,b,c)=1$. Now so certainly $k=4$.

shoki wrote: Find the least value of $k$ such that for all $a,b,c,d \in \mathbb{R}$ the inequality \[ \begin{array} c \sqrt{(a^2+1)(b^2+1)(c^2+1)} +\sqrt{(b^2+1)(c^2+1)(d^2+1)} +\sqrt{(c^2+1)(d^2+1)(a^2+1)} +\sqrt{(d^2+1)(a^2+1)(b^2+1)} \\ \ \\ \ge 2( ab+bc+cd+da+ac+bd)-k \end{array}\]holds. Using Complex Numbers, we get the following nice identity: \[(x^2+1)(y^2+1)(z^2+1)=(xy+yz+zx-1)^2+(x+y+z-xyz)^2 \geq (xy+yz+zx-1)^2\]Then we conclude that: \[ \sum \sqrt{(a^2+1)(b^2+1)(c^2+1)} \geq \sum (ab+bc+ac-1) = 2( ab+bc+cd+da+ac+bd)-4\]and we have the equality at $a=b=c=d=\sqrt{3}$, so $k=4$.

shoki wrote: Find the least value of $k$ such that for all $a,b,c,d \in \mathbb{R}$ the inequality \[ \begin{array} c \sqrt{(a^2+1)(b^2+1)(c^2+1)} +\sqrt{(b^2+1)(c^2+1)(d^2+1)} +\sqrt{(c^2+1)(d^2+1)(a^2+1)} +\sqrt{(d^2+1)(a^2+1)(b^2+1)} \\ \ \\ \ge 2( ab+bc+cd+da+ac+bd)-k \end{array}\]holds. I found generalization this problem: Let $a_1,a_2,...,a_n \in \mathbb{R}$ ,$k \in \mathbb{R^*}$ $$\sum_{i=1}^{n}\sqrt{(a_i^2+k^2)(a_{i+1}^2+k^2)(a_{i+2}^2+k^2)}+nk^2\ge(1+k)\sum_{i=1}^{n}a_i a_{i+1}+\sum_{i=1}^{n}a_i a_{i+2}$$

Solved with nukelauncher. The answer is \(k=4\), achieved by \(a=b=c=d=\sqrt3\). It is sufficient to verify that for all real numbers \(a\), \(b\), \(c\), \[\sqrt{(a^2+1)(b^2+1)(c^2+1)}\ge ab+bc+ca-1.\] It is not hard to show that \begin{align*} &(a^2+1)(b^2+1)(c^2+1)-(ab+bc+ca-1)^2\\ &=a^2b^2c^2-2abc(a+b+c)+a^2+b^2+c^2+2ab+2bc+2ca\\ &=a^2b^2c^2-2abc(a+b+c)+(a+b+c)^2\\ &=(abc-a-b-c)^2\ge0, \end{align*}end proof.

TheUltimate123 wrote: Solved with nukelauncher. The answer is \(k=4\), achieved by \(a=b=c=d=\sqrt3\). It is sufficient to verify that for all real numbers \(a\), \(b\), \(c\), \[\sqrt{(a^2+1)(b^2+1)(c^2+1)}\ge ab+bc+ca-1.\] It is not hard to show that \begin{align*} &(a^2+1)(b^2+1)(c^2+1)-(ab+bc+ca-1)^2\\ &=a^2b^2c^2-2abc(a+b+c)+a^2+b^2+c^2+2ab+2bc+2ca\\ &=a^2b^2c^2-2abc(a+b+c)+(a+b+c)^2\\ &=(abc-a-b-c)^2\ge0, \end{align*}end proof. See here