Bài 4:
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{a^2}}=2\)
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2\Rightarrow3\left(\dfrac{a}{b}+\dfrac{b}{c}\right)\ge6\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2-6=-4\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+4\ge-4+4=0\) (đúng)
Hình vẽ không chính xác lắm thông cảm
a) Vì OM song song với AB nên \(\dfrac{OM}{AB}=\dfrac{OD}{BD}\)
Vì OM song song với CD nên \(\dfrac{OM}{CD}=\dfrac{OA}{AC}\)
Vì AB song song với CD nên \(\dfrac{OA}{AC}=\dfrac{OB}{BD}\) nên \(\dfrac{OM}{CD}=\dfrac{OB}{BD}\)
Do đó \(\dfrac{OM}{AB}+\dfrac{OM}{CD}=\dfrac{OD}{BD}+\dfrac{OA}{AC}=\dfrac{OD}{BD}+\dfrac{OB}{BD}=1\)
Hay \(OM\left(\dfrac{1}{AB}+\dfrac{1}{CD}\right)=1\) suy ra \(\dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{1}{OM}\)
Lại có ON song song với CD nên \(\dfrac{ON}{CD}=\dfrac{OB}{BD}\) mà \(\dfrac{OB}{BD}=\dfrac{OM}{CD}\) nên \(\dfrac{ON}{CD}=\dfrac{OM}{CD}\) hay OM = ON = \(\dfrac{1}{2}\)MN
Suy ra \(\dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{1}{OM}=\dfrac{1}{\dfrac{1}{2}MN}=\dfrac{2}{MN}\)
b) Dễ chứng minh SADC = SBDC
Mà SADC = SAOD+SOCD và SBDC = SBOC+SOCD
Suy ra SAOD = SBOC
Lại có \(\dfrac{S_{AOD}}{S_{AOB}}=\dfrac{OD}{OB}\) và \(\dfrac{S_{OCD}}{S_{BOC}}=\dfrac{OD}{OB}\)
Nên \(\dfrac{S_{AOD}}{S_{AOB}}=\dfrac{S_{OCD}}{S_{BOC}}\) \(\Leftrightarrow\) \(S_{AOD}.S_{BOC}=S_{AOB}.S_{OCD}\)
Hay \(S_{AOD}=S_{BOC}=\sqrt{S_{AOB}.S_{OCD}}=\sqrt{a^2.b^2}=ab\)
Khi đó \(S_{ABCD}=S_{AOD}+S_{BOC}+S_{AOB}+S_{OCD}=ab+ab+a^2+b^2=a^2+b^2+2ab=\left(a+b\right)^2\)
