Trước hết , ta khai triển vế trái , sau đó , nhóm các hạng tử .

\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+b^2d^2+2abcd+a^2d^2+b^2c^2-2abcd\)

\(=\left(a^2c^2+a^2d^2\right)+\left(b^2c^2+b^2d^2\right)\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

Vậy \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\left(ĐPCM\right)\)

(ac+bd)^2=\(^{a^2c^2+2abcd+b^2d^2}\) 

\(\left(ad-bc\right)^2=a^2d^2-2abcd+b^2c^2\)

\(\Rightarrow\left(ac+bd\right)^2-\left(ad-bc\right)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2\) =vp(dpcm)

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Trước hết theo BĐT Schur bậc 3 ta có:

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)^2}{a^2c^2+2ab^2c}\)

\(\Rightarrow P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)

Áp dụng (1):

\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)

=> (a+d)² - (b+c)²= (a-d)² - (c-b)²

=> a²+ d²+ 2ad - b²- c²- 2bc=a² + d² - 2ad - c²-b²+2bc

Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)

1) a²+b²+c²+3=2(a+b+c) =>(a-1)²+(b-1)²+(c-1)²=0

=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm

2) (a+b+c)²=3(ab+bc+ac) =>(a-b)²+(b-c)²+(c-a)²=0

=>a-b=b-c=c-a=0 =>a=b=c

\(S=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

\(S=\left(a^2+ab+bc+ac\right)\left(b^2+ab+bc+ac\right)\left(c^2+ab+bc+ac\right)\)

\(S=\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

là số chính phương (đpcm)

Có : \(a^2+1=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự : \(b^2+1=\left(a+b\right)\left(b+c\right)\)và    \(c^2+1=\left(a+c\right)\left(b+c\right)\)

Suy ra : \(S=\left(a+b\right)\left(a+c\right).\left(a+b\right)\left(b+c\right).\left(a+c\right)\left(b+c\right)\)

\(\Leftrightarrow S=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\)là số chính phương \(\forall\)a ,b ,c nguyên !

với ab+bc+ca=1, ta có 

\(a^2+1=a^2+ab+bc+ca=\left(a^2+ab\right)+\left(bc+ca\right)\)\(=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)

tương tự tra có \(b^2+1=\left(a+b\right)\left(b+c\right)\)

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

=> S=\(\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\) 

mà a,b, c là các số nguyên => \(\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\) là số chính phương 

=> S là số chính phương (ĐPCM)

ta có \(\hept{\begin{cases}a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+c\right)\\b^2+1=b^2+ab+bc+ca=b\left(a+b\right)+a\left(b+c\right)=\left(b+c\right)\left(b+a\right)\\c^2+1=c^2+ab+bc+ca=c\left(b+c\right)+a\left(b+c\right)=\left(c+a\right)\left(b+c\right)\end{cases}}\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(c+a\right)\left(b+c\right)\)\(=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

=> đpcm