I think the original RHS is $\sqrt[n]{1-(x_1-x_n)^2}$.

sqing wrote: Let $0\leq x_1\leq x_2\leq \cdots \leq x_n\leq 1 $ $(n\geq 2).$ Prove that $$\sqrt[n]{x_1x_2 \cdots x_n}+ \sqrt[n]{(1-x_1)(1-x_2)\cdots (1-x_n)}\leq \sqrt[n]{1-(x_1- x_n)^2}.$$ Nice Holder!

Yes, for two sequences: $$[x_1x_n,x_2,x_3,...,x_{n-1}],[(1-x_1)(1-x_n),1-x_2,1-x_3,...,1-x_{n-1}]$$

And of course the finishing step: $(1-x_1)(1-x_n)\le 1-(x_1-x_n)^2\iff x_1(1-x_1)+x_n(1-x_n)+x_1x_n\ge 0$ Putnam 2003 A2 \[(a_1a_2 \cdots a_n)^{1/n}+ (b_1b_2 \cdots b_n)^{1/n} \le ((a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n}\]

Putnam 2003 A2: Let $a_1, a_2, \cdots , a_n$ and $b_1, b_2,\cdots, b_n$ be nonnegative real numbers. Show that \[\sqrt[n]{a_1a_2 \cdots a_n}+\sqrt[n]{b_1b_2 \cdots b_n} \leq \sqrt[n]{(a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n)}.\]

sqing wrote:

Putnam 2003 A2: Let $a_1, a_2, \cdots , a_n$ and $b_1, b_2,\cdots, b_n$ be nonnegative real numbers. Show that \[\sqrt[n]{a_1a_2 \cdots a_n}+\sqrt[n]{b_1b_2 \cdots b_n} \leq \sqrt[n]{(a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n)}.\] Yes, for $[x_1,x_n,x_2,x_3,\cdots,x_{n-1}],[1-x_n,1-x_1,1-x_2,1-x_3,\cdots,1-x_{n-1}]$.

Sniped, but still posting. WolfusA wrote: Yes, for two sequences: $$[x_1x_n,x_2,x_3,...,x_{n-1}],[(1-x_1)(1-x_n),1-x_2,1-x_3,...,1-x_{n-1}]$$ You are wrong, because you get Holder for $n-1$ not $n$. Here is a correct solution. Apply Holder to get $$(x_1+(1-x_n))(x_2+(1-x_2))\dots (x_{n-1}+(1-x_{n-1}))(x_n+(1-x_1))\geq \left(\sqrt[n]{x_1x_2\dots x_n}+\sqrt{(1-x_1)(1-x_2)\dots(1-x_n)} \right)^n$$Wich is equivalent to $$(1-(x_n-x_1))(1+(x_n-x_1))\geq \left(\sqrt[n]{x_1x_2\dots x_n}+\sqrt{(1-x_1)(1-x_2)\dots(1-x_n)} \right)^n$$$\blacksquare$

Nice, Ez and a beautiful problem. Notice that we are asked to prove \[ (1 - x_1 + x_n)( 1 - x_n + x_1) \ge \left( \sqrt[n]{x_1 x_2 x_3 \dots x_n} + \sqrt[n]{(1-x_1)(1-x_2)\dots (1-x_n)} \right)^n \]Apply Holder the following way, \[ (1 - x_1 + x_n )(1 - x_n + x_1) \prod_{k = 2}^{n-1} (1 - x_k + x_k) \ge \left( \sqrt[n]{x_1 x_2 x_3 \dots x_n} + \sqrt[n]{(1-x_1)(1-x_2)\dots(1-x_n)} \right)^n \]

But what if all of Xi be 1???

Eliti wrote: But what if all of Xi be 1??? We'll get $1\geq1.$