Sửa đề: a,b,c,d>0

C/m: \(\left(\frac{a+b}{2}+\frac{c+d}{2}\right)^2\ge\left(a+c\right)\left(c+d\right)\)

Áp dụng BĐT AM-GM ta có:

\(\left(\frac{a+b}{2}+\frac{c+d}{2}\right)^2=\left[\frac{\left(a+c\right)+\left(b+d\right)}{2}\right]^2\ge\left[\frac{2.\sqrt{\left(a+c\right)\left(b+d\right)}}{2}\right]^2=\left(a+c\right)\left(b+d\right)\)

Dấu " = " xảy ra <=> a+c=b+d

Sorry, đề bài thiếu: a,b,c,d là số dương

Ta có : \(\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow a^2+b^2-2ab\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge2ab+a^2+b^2=\left(a+b\right)^2\left(1\right)\)

Chia cả 2 vế của \(\left(1\right)\)cho 4 , ta được :

\(\frac{a^2+b^2}{2}\ge\frac{\left(a+b\right)^2}{4}=\left(\frac{a+b}{2}\right)^2\)

\(\Rightarrowđpcm\)

Xét hiệu

\(\frac{a^2+b^2+c^2}{3}-\left(\frac{a+b+c}{3}\right)^2\)

\(=\frac{a^2+b^2+c^2}{3}-\frac{\left(a+b+c\right)^2}{9}\)

\(=\frac{3\left(a^2+b^2+c^2\right)}{9}-\frac{a^2+b^2+c^2+2ab+2bc+2ac}{9}\)

\(=\frac{1}{9}\left[3\left(a^2+b^2+c^2\right)-a^2-b^2-c^2-2ab-2bc-2ac\right]\)

\(=\frac{1}{9}\left(3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ac\right)\)

\(=\frac{1}{9}\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\)

\(=\frac{1}{9}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\right]\)

\(=\frac{1}{9}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\) \(\ge0\)

Vậy \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)

Dấu "=" xảy ra <=> a=b=c

\(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\Leftrightarrow\frac{a^3+b^2+c^2}{3}\ge\frac{\left(a+b+c\right)^2}{9}\)

\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\Leftrightarrow3a^2+3b^2+3c^2\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ac\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Các phép biến đổi là tương đương suy ra đpcm. \("="\Leftrightarrow a=b=c\)

Kiểm tra lại đề nhé! 

Em thử cho a = b = c xem sao?

sửa số 2 thành số 8 nha

kiem tra de ban oi