Question

C/m bất đẳng thức: \(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt[]{ab}+\sqrt[]{cd}\) ( với a, b, c, d >0)

Answer 1

\(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\)

<=> \(\left(a+c\right)\left(b+d\right)\ge ab+2\sqrt{abcd}+cd\)  (bình phương hai vế)

<=> \(ab+ad+bc+cd\ge ab+2\sqrt{abcd}+cd\)

<=>\(ad-2\sqrt{abcd}+bc\ge0\)

<=> \(\left(\sqrt{ad}-\sqrt{bc}\right)^2\ge0\)  luôn luôn đúng với a,b,c,d>0

=>đpcm

 

Answer 2

Áp dụng BĐT bu nhi a cốp-xki cho 4 số dương ,ta có:

\(\left(\sqrt{a}^2+\sqrt{c}^2\right)\left(\sqrt{b}^2+\sqrt{d}^2\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)

hay \(\left(a+c\right)\left(b+d\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)

→\(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\)(đfcm)