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

Ta có:

\(VT=\left[\dfrac{16a-a^2-\left(3+2a\right)\left(a+2\right)-\left(2-3a\right)\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}\right]:\dfrac{a-1}{a^3+4a^2+4a}\)

\(=\dfrac{16a-a^2-3a-6-2a^2-4a-2a+4+3a^2-6a}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}\)

\(=\dfrac{a-2}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}=\dfrac{a\left(a+2\right)}{a-1}\left(a\ne\pm2;a\ne1\right)\)

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

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

\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)

b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

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

\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)

Vậy ta có đpcm 

b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)

Vậy ta có đpcm 

a) Ta có: \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}\)

\(=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)

\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1\)

b) Ta có: \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+\sqrt{a}+a\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}\right)^2:\left(1+\sqrt{a}\right)^2=1\)

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

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

áp dụng bất đẳng thức cô si cho:

*a+b≥\(2\sqrt{ab}\)

*b+c≥\(2\sqrt{bc}\)

*c+a≥\(2\sqrt{ca}\)

➩2(a+b+c)≥2(\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\))

➩ĐPCM

Ta có:

\(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\Leftrightarrow2a+2b+2c\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ca}+a\right)\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt[]{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)

(luôn đúng với mọi a,b,c không âm)

Dấu bằng xảy ra \(\Leftrightarrow a=b=c\)

Đây nhé! Tích giúp c nha

a/ Chứng minh:

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

\(=x^2+bx+ax+ab\)

\(=x^2+\left(ax+bx\right)+ab\)

\(=x^2+x\left(a+b\right)+ab=VP\) (đpcm)

b/ Chứng minh:

\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)

\(=x^3+cx^2+ax^2+acx+bx^2+bcx+abx+abc\)

\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+bcx+acx\right)+abc\)

\(=x^3+x^2\left(a+b+c\right)+x\left(ab+bc+ac\right)+abc=VP\) (đpcm)