Question

CM BĐT sau

a/ \(\left(a^2-b^2\right)\left(c^2-d^2\right)\le\left(ac-bd\right)^2\) \(\forall a,b,c,d\)

b/ \(\left(1+a^2\right)\left(1+b^2\right)\ge\left(1+ab\right)^2\) \(\forall a,b\)

c/ \(a^2+b^2+1\ge ab+a+b\) \(\forall a,b\)

Answer 1

c) theo bđt cauchy ta có

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+1\ge2b\\a^2+1\ge2a\end{matrix}\right.\)

cộng hết lại rút 2 đi \(\Rightarrowđpcm\)

Answer 2

b)theo bđt bunhiacopxki ta có

\(\left(1^2+a^2\right)\left(1^2+b^2\right)\ge\left(1+ab\right)^2\)

\(\Rightarrowđpcm\)

Answer 3

theo bđt cauchy ta có

\(-\left(a^2d^2+b^2c^2\right)\le-2abcd\)

\(\Leftrightarrow a^2c^2-a^2d^2+b^2d^2-b^2c^2\le a^2c^2-2abcd+b^2d^2\)

\(\Leftrightarrow a^2(c^2-d^2)-b^2(c^2-d^2)\le a^2c^2-2abcd+b^2d^2\)

\(\Leftrightarrow(c^2-d^2)\left(a^2-b^2\right)\le(ac-bd)^2\)

\(\Rightarrowđpcm\)