\(a^3+a^3+b^3\ge3\sqrt[3]{a^6b^3}=3a^2b\)

\(b^3+b^3+a^3\ge3b^2a\)

\(\Rightarrow3\left(a^3+b^3\right)\ge3\left(a^2b+b^2a\right)\Leftrightarrow\left(a^3+b^3\right)\ge\left(a^2b+b^2a\right)\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)

a) A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)

=> A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)

=> A = \(\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

=> A + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

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

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

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

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

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

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

\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(A=1-\frac{1}{n+1}\)

a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

           \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

           \(A=1-\frac{1}{n+1}\)

           \(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)

           \(A=\frac{n}{n+1}\)

Học tốt nha^^

2017 = 2016 + 1 = x + 1

suy ra 2017x¹⁵ = x¹⁶ + x¹⁵

2017x¹⁴ = x¹⁵ + x¹⁴

.... 

từ đó ta dễ tính ra A

\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(A=x^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left[\left(y-z\right)+\left(z-x\right)\right]\)

\(A=x^4\left(y-z\right)-z^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left(z-x\right)\)

\(A=\left(y-z\right)\left(x^4-z^4\right)+\left(z-x\right)\left(y^4-z^4\right)\)

\(A=\left(y-z\right)\left(x-z\right)\left(x+z\right)\left(x^2+z^2\right)-\left(x-z\right)\left(y-z\right)\left(y+z\right)\left(y^2+z^2\right)\)

\(A=\left(y-z\right)\left(x-z\right)\left(x^3+xz^2+x^2z+z^3-y^3-yz^2-y^2z-z^3\right)\)

\(A=\left(y-z\right)\left(x-z\right)\left(x-y\right)\left(x^2+xy+y^2+z^2+zx+yz\right)\)

\(A=\frac{1}{2}\left(x-y\right)\left(y-z\right)\left(x-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)

Vì \(x>y>z\Rightarrow A>0\)