+\(\frac{a}{b}=1\Leftrightarrow a=b\Leftrightarrow\frac{a}{b}=\frac{a+2016}{b+2016}\)

+\(\frac{a}{b}>1\Leftrightarrow a>b\Leftrightarrow\frac{a}{b}-1=\frac{a-b}{b}>\frac{a-b}{b+2016}=\frac{a+2016}{b+2016}-1\)=> \(\frac{a}{b}>\frac{a+2016}{b+2016}\)

+\(\frac{a}{b}< 1\Leftrightarrow a< b\Leftrightarrow1-\frac{a}{b}=\frac{b-a}{b}>\frac{b-a}{b+2016}=1-\frac{a+2016}{b+2016}\)=>\(\frac{a}{b}< \frac{a+2016}{b+2016}\)

A+2016/B+2016=A/B+2016/2016=A/B+1

=)A/B<A/B+1

=)A/B<A+2016/B+2016

\(\frac{a}{b}\)<\(\frac{a+2016}{b+2016}\)

Xét hiệu:

\(H=\frac{a}{b}-\frac{a+2016}{b+2016}=\frac{a\cdot\left(b+2016\right)-\left(a+2016\right)\cdot b}{b\left(b+2016\right)}=\frac{2016\cdot\left(a-b\right)}{b\left(b+2016\right)}.\)

tùy H>0 hay H<0 mà ta biết được kq của sự so sánh.

Ta có: \(\frac{a}{b+2016}< \frac{a}{b}\) và \(\frac{2016}{b+2016}< \frac{a}{b}\)

=>  \(\frac{a}{b+2016}+\frac{2016}{b+2016}< \frac{a}{b}\)

hay \(\frac{a+2016}{b+2016}< \frac{a}{b}\)

n

nếu a>b hay a/b > 1 ta có 2016a > 2016b 

                                => 2016a + ab > 2016b + ab 

                               => a ( 2016 + b) > b ( 2016 + a )

                               => a/b > a+2016/b+2016 

tương tự với 2 trường hợp

 nếu a < b thì a/b < a+2016/b+2016

nếu a = b thì a/b = a+2016/b+2016

Ta có

\(1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(1\Leftrightarrow x^2+\frac{\left(b^2+c^2\right)x^2}{a^2}+y^2+\frac{\left(a^2+c^2\right)y^2}{b^2}+z^2+\frac{\left(a^2+b^2\right)z^2}{c^2}=x^2+y^2+z^2\)

\(\Leftrightarrow\frac{\left(b^2+c^2\right)x^2}{a^2}+\frac{\left(c^2+a^2\right)y^2}{b^2}+\frac{\left(a^2+b^2\right)z^2}{c^2}=0\)

Ta thấy rằng cả 3 phân số đó đều \(\ge0\)nên tổng 3 phân số sẽ \(\ge0\)

Dấu = xảy ra khi x = y = z = 0

Với x = y = z = 0 thì

\(\frac{x^{2016}}{a^{2016}}+\frac{y^{2016}}{b^{2016}}+\frac{z^{2016}}{c^{2016}}=\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\Leftrightarrow\frac{0}{a^{2016}}+\frac{0}{b^{2016}}+\frac{0}{c^{2016}}=\frac{0+0+0}{a^{2016}+b^{2016}+c^{2016}}\)

\(\Leftrightarrow0=0\)(đúng)

\(\Rightarrow\)ĐPCM

Ta có:

\(\frac{a}{b}\)= \(\frac{a\left(b+2016\right)}{b\left(b+2016\right)}\)=\(\frac{ab+2016a}{b\left(b+2016\right)}\)

\(\frac{a+2016}{b+2016}\)=\(\frac{\left(a+2016\right)b}{\left(b+2016\right)b}\)=\(\frac{ab+2016b}{b\left(b+2016\right)}\)

Vì b > 0 nên mẫu số của hai phân số trên dương. Ta so sánh tử số.

* Nếu a < b => ab+2016a < ab+2016b

=> \(\frac{a}{b}\)<\(\frac{a+2016}{b+2016}\)

* Nếu a = b => ab+2016a = ab+2016b

=> \(\frac{a}{b}\)=\(\frac{a+2016}{b+2016}\)

* Nếu a > b => ab+2016a > ab+2016b

=> \(\frac{a}{b}\)>\(\frac{a+2016}{b+2016}\)

Giả sử a/b = 1/3 còn phân số kia là 2017/2019

quy đồng 1/3 = 2017/6051

Vì 6051>2019 nên 2017/2019 > 2017/6051 và 2017/2019>1/3

Vậy \(\frac{a}{b}< \frac{a+2016}{b+2016}\)

Ta có : \(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)

           \(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)

Vì \(\frac{3}{2016^{2016}-1}>\frac{3}{2016^{2016}-3}\)

\(\Rightarrow1+\frac{3}{2016^{2016}-1}>1+\frac{3}{2016^{2016}-3}\)

\(\Rightarrow A>B\)