Ta có :

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)⇒\(\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

\(\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)\(\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+b^2\right)\left(1+a^2\right)\left(1+ab\right)}\ge0\).vì ab≥0 nên sua ra đpcm

a)TXĐ:\(x-1>0\Rightarrow x>1\)

áp dụng công thức :\(v=v_0+at\Rightarrow v_0=-2a\)

và\(v^2-v^{2_0}=2as\Rightarrow v^{2_0}=2as\Rightarrow\left(2a\right)^2=3.6a\)

\(\Rightarrow a=\dfrac{3.6}{4}m/s^2\)

ADCT:\(F_{hl}=ma\Rightarrow F=500\cdot\dfrac{3.6}{4}=450N\)