The Lagrange multipliers method easily establishes that $\min F_n = -\frac {1} {n}$ for $n=3,4,5$, with equality when all variables are equal.

mavropnevma, for $n$ ?

$F_3=-1+2(x_1^2+x_2^2+x_3^2)\ge -1+\frac23=-\frac 13$ $F_4=-1+\frac 32\left[(x_1+x_3)^2+(x_2+x_4)^2\right]+\frac 12\left[(x_1-x_3)^2+(x_2-x_4)^2\right]\ge$ $-1+\frac32\left[(x_1+x_3)^2+(x_2+x_4)^2\right]\ge-\frac 14$ $F_5=-1+\left[(x_1+x_3)^2+(x_1+x_4)^2+(x_2+x_4)^2+(x_2+x_5)^2+(x_3+x_5)^2\right]\ge$ $ -1+\frac45=-\frac 15$

$F_3=(x_1+x_2+x_3)^2-4(x_1x_2+x_2x_3+x_3x_1)$ $\ge (x_1+x_2+x_3)^2-\frac 43(x_1+x_2+x_3)^2=-\frac 13$.

mavropnevma wrote: The Lagrange multipliers method easily establishes that $\min F_n = -\frac {1} {n}$ for $n=3,4,5$, with equality when all variables are equal. Can one tell me what do you mean by 'Lagrange multipliers method'??..

sayantanchakraborty wrote: Can one tell me what do you mean by 'Lagrange multipliers method'??.. https://en.wikipedia.org/wiki/Lagrange_multiplier

Thanks... Min $ab+bc+ac=1/3$ and min $F_n=1–4/3=–1/3$

This problem is natural for Lagrange murderpliers So first of all we must estabilsh that me may proceed with the murderpliers. So now we will define an closed set $\overline{U} = [0,1]^n$ We may define a constraint set $G = \{x\in \overline{U} \mid g(x)=1 \}$, where $g(x_1,x_2,...,x_n) = x_1+x_2+...x_n$ So define $f(x_1,x_2,...,x_n) = F_n$ Now as we see $\triangledown g(x) \neq 0$, where $x \in \overline{U}$ Thus we have the following: $$\triangledown f(x) = \lambda \triangledown g(x)$$Solving this we get that $x_1=x_2=....=x_n$ So now we get that the minimum is achieved when every $x_i$ is $\frac{1}{n}$. Putting this in and solving it we get a true statement Thus $F_n \geqslant -\frac{1}{n} $ a)$F_3 \geqslant -\frac{1}{3}$ b)$F_4 \geqslant -\frac{1}{4}$ c)$F_5 \geqslant -\frac{1}{5}$ ............