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

Solution 1:

$a^2+b^2+c^2+d^2=1$ => $a,b,c,d\in[0,1]$. 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 =>$xyzw\geq1$

Solution 2:

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.

Solution 3:

Observe that $f(x)=\log\left(\frac1x-1\right)$ is concave for $0<x<1$ (its derivative is $\frac1{x(1-x)}$).

Taking $\log$, the inequality is equivalent to $$\sum\log\left(\frac1a-1\right)\geq0.$$ By Jensen's inequality, $$\sum\log\left(\frac1a-1\right)\geq4\log\left(\frac4{a+b+c+d}-1\right).$$ By Cauchy-Schwarz/Power Mean/Hölder, $a+b+c+d\leq2\sqrt{a^2+b^2+c^2+d^2}=2$, hence $$LHS\geq4\log\left(\frac42-1\right)=0,$$ as was to be shown.