` a,b,c,d ` satisfy ` ad -bc =1` . Prove that ` a^2 +b^2 +c^2+d^2 + ac+bd \ge \sqrt(3)`

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1. Đáp án:

   `+) (a^2 + b^2)(c^2 + d^2) = (ad – bc)^2 + (ac + bd)^2`  (hiển nhiên)

                                               `= 1 + (ac + bd)^2`

    Đặt `Z = a^2 + b^2 + c^2 + d^2 + ac + bd`

   `(Cosi) -> Z ≥ 2\sqrt{(a^2 + b^2)(c^2 + d^2)} + ac + bd = 2\sqrt{1 + (ac + bd)^2} + ac + bd (1)`

   Đặt `ac + bd = x`

    `(1) <=> Z ≥ 2\sqrt{1 + x^2} + x`

   `<=> Z^2 ≥ x^2 + 4(1 + x^2) + 4x\sqrt{1 + x^2} = (x^2 + 1) + 2.\sqrt{x^2 + 1}.2x + 4x^2 + 3`

   `= (\sqrt{x^2 + 1} + 2x)^2 + 3 ≥ 3 (1)`

   Măt khác : `Z = a^2 + b^2 + c^2 + d^2 + ac + bd`

   `= 1/2 (a^2 + b^2 + c^2 + d^2) + 1/2 (a^2 + 2ac + c^2) + 1/2 (b^2 + 2bc + d^2)`

   `= 1/2 (a^2 + b^2 + c^2 + d^2) + 1/2 (a + c)^2 + 1/2 (b + d)^2 ≥ 0 (2)`

   Từ `(1)(2) -> Z ≥ \sqrt{3}`

   Giải thích các bước giải:

    

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2. Giải thích các bước giải:

   Ta có:

   $S=a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}$

   $\to S=a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}(ad-bc)$

   $\to S=a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}ad+\sqrt{3}bc$

   $\to S=a^2+(ac-\sqrt{3}ad)+b^2+c^2+d^2+bd+\sqrt{3}bc$

   $\to S=a^2+a(c-\sqrt{3}d)+b^2+c^2+d^2+bd+\sqrt{3}bc$

   $\to S=a^2+2a\cdot \dfrac{c-\sqrt{3}d}{2}+( \dfrac{c-\sqrt{3}d}{2})^2-( \dfrac{c-\sqrt{3}d}{2})^2+b^2+c^2+d^2+bd+\sqrt{3}bc$

   $\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+3c^2+2\sqrt{3}cd+4\sqrt{3}cb+d^2+4db}{4}$

   $\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+4\sqrt{3}cb+4db+d^2+3c^2+2\sqrt{3}cd}{4}$

   $\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+4b(\sqrt{3}c+d)+3c^2+2\sqrt{3}cd+d^2}{4}$

   $\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{4b^2+4b(\sqrt{3}c+d)+(\sqrt{3}c+d)^2}{4}$

   $\to S=(a+ \dfrac{c-\sqrt{3}d}{2})^2+\dfrac{(2b+\sqrt{3}c+d)^2}{4}$

   $\to S\ge 0$

   $\to a^2+b^2+c^2+d^2+ac+bd-\sqrt{3}\ge 0$

   $\to a^2+b^2+c^2+d^2+ac+bd\ge \sqrt{3}$

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