Giải thích các bước giải:
Ta có:
$S=a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}$
$\to S=a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}(ad-bc)$
$\to S=a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}ad+\sqrt{3}bc$
$\to S=a^2+(ac-\sqrt{3}ad)+b^2+c^2+d^2+bd+\sqrt{3}bc$
$\to S=a^2+a(c-\sqrt{3}d)+b^2+c^2+d^2+bd+\sqrt{3}bc$
$\to S=a^2+2a\cdot \dfrac{c-\sqrt{3}d}{2}+( \dfrac{c-\sqrt{3}d}{2})^2-( \dfrac{c-\sqrt{3}d}{2})^2+b^2+c^2+d^2+bd+\sqrt{3}bc$
$\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+3c^2+2\sqrt{3}cd+4\sqrt{3}cb+d^2+4db}{4}$
$\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+4\sqrt{3}cb+4db+d^2+3c^2+2\sqrt{3}cd}{4}$
$\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+4b(\sqrt{3}c+d)+3c^2+2\sqrt{3}cd+d^2}{4}$
$\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+4b(\sqrt{3}c+d)+(\sqrt{3}c+d)^2}{4}$
$\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{(2b+\sqrt{3}c+d)^2}{4}$
$\to S\ge 0$
$\to a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}\ge 0$
$\to a^2+b^2+c^2+d^2+ac+bd\ge \sqrt{3}$
