Question

cho S=\(a^2+b^2+c^2+d^2+ac+bd\)

trong đó: ad-bc=1

cmr:\(S\ge\sqrt{3}\)

Answer 1

Giải:

\(S=a^2+b^2+c^2+d^2+ac+bd\)

\(\Leftrightarrow S=a^2+b^2+c^2+d^2-2ac+ac+2bd-bd\)

\(\Leftrightarrow S=a^2-2ac+c^2+b^2+2bd+d^2+ac-bd\)

\(\Leftrightarrow S=\left(a^2-2ac+c^2\right)+\left(b^2+2bd+d^2\right)-\left(ac-bd\right)\)

\(\Leftrightarrow S=\left(a-c\right)^2+\left(b+d\right)^2-1\)

\(\Leftrightarrow S\ge-1\)

\(\Leftrightarrow S\ge\sqrt{3}\left(\sqrt{3}>1\right)\)

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