Show that $\sqrt[n]{x_1x_2\dots x_n}+\sqrt[n]{(1-x_1)(1-x_2)\dots(1-x_n)}\leq 1$

Question

Question: Let $x_i\in(0,\frac12]$ for $1\leq i\leq n$. Show that $$\sqrt[n]{x_1x_2\dots x_n}+\sqrt[n]{(1-x_1)(1-x_2)\dots(1-x_n)}\leq 1$$ without using optmization techniques in calculus.

My try: For $n=1$, it is trivially true. For $n=2$, let $x_1=u^2,x_2=v^2$. Then, $$\sqrt{1-u^2-v^2+u^2v^2}\leq 1-uv$$ then after squaring $$0\leq (u-v)^2.$$ At $n=3$, I am stuck. Is there an easier method to prove this inequality? Also, what is the lower bound of this sum? Thanks in advance.

Answer 1

By using AM-GM inequality, $$\sqrt[n]{x_1x_2\dots x_n}+\sqrt[n]{(1-x_1)(1-x_2)\dots(1-x_n)}\leq \frac{x_1+\ldots+x_n}{n} + \frac{n-(x_1+\ldots+x_n)}{n} = 1$$

Answer 2

Because by Holder: $$1=\sqrt[n]{\prod_{i=1}^n(x_i+1-x_i)}\geq\sqrt[n]{\prod_{i=1}^nx_i}+\sqrt[n]{\prod_{i=1}^n(1-x_i)}.$$