a) \(\dfrac{a^2+a+1}{a^2-a+1}=\dfrac{\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Thấy tử và mẫu của phân số đều lớn hơn 0 => \(\dfrac{a^2+a+1}{a^2-a+1}>0\)

b)\(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2a+1\right)+\left(c^2-2a+1\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (luôn đúng với mọi a,b,c)

Dấu = xra khi a=b=c=1

b)

\(a^2-2a+1+b^2-2b+1+c^2-2c+1\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) ( Luôn đúng)

Dấu "=" xảy ra khi a=b=c=1

2:

a: =>a^2+2ab+b^2-2a^2-2b^2<=0

=>-(a^2-2ab+b^2)<=0

=>(a-b)^2>=0(luôn đúng)

b; =>a^2+b^2+c^2+2ab+2ac+2bc-3a^2-3b^2-3c^2<=0

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

=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)

1) \(\left(a+b\right)^2\)

\(=\left(a+b\right)\left(a+b\right)\)

\(=a^2+ab+ab+b^2\)

\(=a^2+2ab+b^2\left(dpcm\right)\)

2) \(\left(a-b\right)^3\)

\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)\)

\(=\left(a^2-ab-ab+b^2\right)\left(a-b\right)\)

\(=\left(a^2-2ab+b^2\right)\left(a-b\right)\)

\(=a^3-a^2b-2a^2+2ab^2+ab^2-b^3\)

\(=a^3-3a^2b+3ab^2-b^3\left(dpcm\right)\)

`a)` 

`(a+b)^2`

`=(a+b)(a+b)`

`=a^2+ab+ab+b^2`

`=a^2+2ab+b^2`

`->` ĐPCM

`b)` `(a-b)^3`

`=(a-b)(a-b)(a-b)`

`=(a^2-2ab+b^2)(a-b)`

`=a^3-3a^2b+3ab^2-b^3`

`->` ĐPCM

\(a,VT=-\left(-a+b\right)-\left(b+c\right)-\left(a-c\right)=a-b-b-c-a+c=-2b=VP\)

\(b,VT=-\left(a-b-c\right)+\left(-b+c+a\right)-\left(a+b-c\right)=-a+b+c-b+c+a-a-b+c=-a-b+3c=VP\)

a) Áp dụng BĐT Cosi với ab>0, ta có: 

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2\)(đpcm)

b) Ta có: \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)

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

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

a)Xét \(\left(\dfrac{a+b}{2}\right)^2-\dfrac{a^2+b^2}{2}=\)\(\dfrac{a^2+2ab+b^2-2\left(a^2+b^2\right)}{4}\)\(=\dfrac{-a^2+2ab-b^2}{4}\)\(=\dfrac{-\left(a-b\right)^2}{4}\le0\forall a;b\)

\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\) (bạn ghi sai đề?) 

Dấu = xảy ra <=> a=b

b) \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)-\left(a^8+b^8\right)\left(a^4+b^4\right)\)

\(=a^{12}+a^{10}b^2+a^2b^{10}+b^{12}-\left(a^{12}+a^8b^4+a^4b^8+b^{12}\right)\)

\(=a^2b^2\left(a^8+b^8-a^6b^2-a^2b^6\right)\)

\(=a^2b^2\left(a^2-b^2\right)\left(a^6-b^6\right)=a^2b^2\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) với mọi a,b

=> \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\)

Dấu = xảy ra <=>a=b

(a-b)³=a³-3a²b+3ab²-b³ (1)

-(b-a)³=-(b³-3b²a+3ba²-a³)=-b³+3ab²-3a²b+a³=a³-3a²b+3ab²-b³ (2)

từ (1) và (2) => VT=VP => đpcm.

(-a-b)²=[(-a)+(-b)]²=(-a)²+2.(-a).(-b)+(-b)²=a²+2ab+b²=(a+b)²

=> VT=VP => đpcm.