Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$

Question

Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$ Prove $\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$ I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work.

Answer 1

Using AM-GM we can write $1-a^2-b^2=c^2+d^2\ge2cd$

So the idea right is to prove this $(1-a)(1-b)\ge cd$

or similary Proving this $2(1-a)(1-b) -2cd\ge 0$

$$2(1-a)(1-b) -2cd \ge 2(1-a)(1-b) -1 +a^2+b^2 = (1-a-b)^2\ge 0$$

So $(1-a)(1-b)\ge cd$ is true.Similarly we can show $(1-c)(1-d)\ge ab$. multiply them to obtain the question inequality.

Answer 2

$a^2+b^2+c^2+d^2=1$ => $a,b,c,d\in[0,1]$. If $\displaystyle abcd=0$ so the inequality is true. If $\displaystyle abcd>0$, set : $x=\frac{1-a}{a}, y=\frac{1-b}{b}, z=\frac{1-c}{c}, w=\frac{1-d}{d}$

We have

$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}=1$

And the inequality become $\displaystyle xyzw\geq1$

We will prove that with $x,y,z,w\geq0$ and $xyzw=1$: $\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}\geq1$ (1)

Using Cauchy-Schwarz we have

$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}\geq \frac{1}{(\frac{x}{y}+1)(xy+1)}+\frac{1}{(\frac{y}{x}+1)(xy+1)} +\frac{1}{(\frac{z}{w}+1)(zw+1)}+\frac{1}{(\frac{w}{z}+1)(zw+1)} =\frac{1}{xy+1}+\frac{1}{zw+1} =\frac{1}{xy+1}+\frac{1}{\frac{1}{xy}+1} =\frac{1}{xy+1}+\frac{xy}{xy+1}=1$ Suppose that $xyzw<1$. Set $t=\frac{1}{xyz}$ so $xyzt=1$ and $t>w$. Using (1), we have $1\le\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2} <\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}=1$ So the suppose is false =>$xyzw\geq1$