<s>[INST] How to prove $a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$? We have $ad-bc=1$ . prove
$$a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$$
To solve it, I multiplied the inequality by $2$ and added $2(ad-bc)=2$ to it :
$$2a^2+2b^2+2c^2+2d^2+2ac+2bd+2ad-2bc\ge2+2\sqrt3$$
$$(a+c)^2+(b+d)^2+(a+d)^2+(b-c)^2\ge 2+2\sqrt3$$
From here I don't know how to prove the minimum value of sum of these squares is $2+2\sqrt3$ .
 [/INST] by AM-GM $$a^2+b^2+c^2+d^2\ge 2\sqrt{(a^2+b^2)(c^2+d^2)}=2\sqrt{{(ac+bd)}^2+{(ad-bc)}^2}$$ $$=2\sqrt{{(ac+bd)}^2+1}$$ Let $ac+bd=t$ we have to prove $$2\sqrt{t^2+1}+t\ge \sqrt{3}$$ Can you end it now?

 we have to prove $$2\sqrt{t^2+1}\ge \sqrt{3}-t$$ if $\sqrt{3}-t\le 0$ the inequality is obvious other wise squaring both sides of inequality it suffices to prove $$4(t^2+1)\ge t^2-2\sqrt{3}t+3\iff {(\sqrt{3}t+1)}^2\ge 0$$ which is true

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