how to prove this inequality $(x-1)^2+(y-1)^2+(z-1)^2+(u-1)^2+(w-1)^2+(2v+1)^2\ge 4$

Question

let $a,b,c,d$ be real numbers, and such $$(a+c)(b+d)=\sqrt{2}(ac-2bd-1)$$ show that $$(ab-1)^2+(bc-1)^2+(cd-1)^2+(da-1)^2+(ac-1)^2+(2bd+1)^2\ge 4$$

My try: let $x=ab,y=bc,z=cd,w=da,u=ac,v=bd$,then have $$x+y+z+w=\sqrt{2}u-2\sqrt{2}v-\sqrt{2}$$ we must prove $$(x-1)^2+(y-1)^2+(z-1)^2+(u-1)^2+(w-1)^2+(2v+1)^2\ge 4$$

But I use Cauchy-Schwarz inequality we have $$[(x-1)^2+(y-1)^2+(z-1)^2+(u-1)^2+(w-1)^2+(2v+1)^2][1+1+1+2+1+2]\ge (x-1+y-1+z-1+w-1+\sqrt{2}u-\sqrt{2}+2\sqrt{2}v+\sqrt{2})^2=(4+\sqrt{2})^2$$ so we have $$(x-1)^2+(y-1)^2+(z-1)^2+(u-1)^2+(w-1)^2+(2v+1)^2\ge\dfrac{(4+\sqrt{2})^2}{8}=\dfrac{9+4\sqrt{2}}{4}$$ and I can't prove the inequality left hand side is big than $4$

Answer 1

By C-S $$\sum_{cyc}(ab-1)^2+(2bd+1)^2+(ac-1)^2=$$ $$=6+\sum_{cyc}a^2b^2-2(a+c)(b+d)+a^2c^2-2ac+4b^2d^2+4bd=$$ $$=5-2(a+c)(b+d)+\sum_{cyc}a^2b^2+4abcd+(ac-2bd-1)^2=$$ $$=(ab+cd)^2+(bc+ad)^2+\frac{1}{2}(a+c)^2(b+d)^2+5-2(a+c)(b+d)=$$ $$=\frac{1}{2}((a+c)(b+d)+2)^2+(ab+cd)^2+(bc+ad)^2-4(a+c)(b+d)+3=$$ $$=(ab+cd-2)^2+(bc+ad-2)^2+\frac{1}{2}((a+c)(b+d)+2)^2-5=$$ $$=\frac{1}{4}(1+1+2)((ab+cd-2)^2+(bc+ad-2)^2+\frac{1}{2}((a+c)(b+d)+2)^2)-5\geq$$ $$\geq\frac{1}{4}\left(-ab-cd+2-bc-ad+2+(a+c)(b+d)+2\right)^2-5=4.$$

Answer 2

Since $(ac-1)^2 + (2bd+1)^2 = (ac - 2bd-1)^2 + 4abcd + 1 = \frac{(a+c)^2(b+d)^2}{2} + 4abcd + 1$,
the inequality becomes $$(ab-1)^2+(bc-1)^2+(cd-1)^2+(ad-1)^2 + \frac{(a+c)^2(b+d)^2}{2} + 4abcd + 1 \ge 4$$ or $$\frac{1}{2}(pa^2 + qa + r) \ge 0$$ where \begin{align} p &= 3 b^2+2 b d+3 d^2, \\ q &= 2 b^2 c+12 b c d+2 c d^2-4 b-4 d, \\ r &= 3 b^2 c^2+2 b c^2 d+3 c^2 d^2-4 b c-4 c d+2. \end{align} Clearly $p \ge 0$.
Also, $r = (bc - cd)^2 + 2(bc + cd)^2 - 4(bc + cd) + 2 = (bc - cd)^2 + 2(bc + cd - 1)^2 \ge 0$,
and $4pr - q^2 = 8(b-d)^2(2bc+2cd-1)^2 \ge 0$. We are done.