Ta có :

\(\dfrac{a}{b}< \dfrac{c}{d}\)

\(\Rightarrow\dfrac{a}{b}-\dfrac{c}{d}< 0\)

\(\Rightarrow\dfrac{ad-bc}{bd}< 0\)

Mà \(bd>0\) (do b,d dương)

\(\Rightarrow\left\{{}\begin{matrix}ad-bc< 0\\bd>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}ad< bc\\bd>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{bd}{ad}>\dfrac{bd}{bc}\)

\(\Rightarrow\dfrac{b}{a}>\dfrac{d}{c}\)

\(\rightarrowđpcm\)

Xét \(( a^2 + b^2 + c^2 + d^2 ) - ( a + b + c + d)\)

\(= a(a -1) + b( b -1) + c( c – 1) + d( d – 1)\)

Vì a là số nguyên dương nên $a$, $(a – 1)$ là hai số tự nhiên liên tiếp

\(\Rightarrow a-1⋮2\)

Tương tự ta có $b(b-1)$; $c(c-1)$; $d(d-1)$ đều chia hết cho 2

=> $a(a -1) + b( b -1) + c( c – 1) + d( d – 1)$ là số chẵn

Lại có \(a^2 + c^2 = b^2 + d^2=> a^2 + b^2 + c^2 + d^2 = 2( b^2 + d^2)\) là số chẵn.

Do đó $a + b + c + d$ là số chẵn mà $a + b + c + d > 2$ (Do \(a,b,c,d\in N^{sao}\))

\(\Rightarrow\) $a + b + c + d$ là hợp số.

Ta có \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)

> \(\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1\)(1)

Tương tự ta chứng minh được \(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{a+d}>1\)(2)

mà \(\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{b}{b+c}+\frac{d}{c+d}+\frac{c}{c+d}+\frac{a}{a+d}+\frac{d}{a+d}=4\)(3)

Từ (1) (2) (3) => \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\left(a;b;c;d\inℕ\right)\)

a)

\(a^2+b^2+c^2+d^2+m^2-a(b+c+d+m)\)

\(=\frac{4a^2+4b^2+4c^2+4d^2+4m^2-4a(b+c+d+m)}{4}\)

\(=\frac{(a^2+4b^2-4ab)+(a^2+4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4m^2-4am)}{4}\)

\(=\frac{(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2m)^2}{4}\geq 0\) (đpcm)

Dấu "=" xảy ra khi \(a=2b=2c=2d=2m\)

b)

Xét hiệu

\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}=\frac{x+y}{xy}-\frac{4}{x+y}=\frac{(x+y)^2-4xy}{xy(x+y)}\)

\(=\frac{x^2+y^2-2xy}{xy(x+y)}=\frac{(x-y)^2}{xy(x+y)}\geq 0, \forall x,y>0\)

\(\Rightarrow \frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\) (đpcm)

Dấu "=" xảy ra khi $x=y$

c)

Xét hiệu:

\((a^2+c^2)(b^2+d^2)-(ab+cd)^2\)

\(=(a^2b^2+a^2d^2+c^2b^2+c^2d^2)-(a^2b^2+2abcd+c^2d^2)\)

\(=a^2d^2-2abcd+b^2c^2=(ad-bc)^2\geq 0\)

\(\Rightarrow (a^2+c^2)(b^2+d^2)\geq (ab+cd)^2\) (đpcm)

Dấu "=" xảy ra khi \(ad=bc\)

d)

Xét hiệu:

\(a^2+b^2-(a+b-\frac{1}{2})=a^2+b^2-a-b+\frac{1}{2}\)

\(=(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})\)

\(=(a-\frac{1}{2})^2+(b-\frac{1}{2})^2\geq 0\)

\(\Rightarrow a^2+b^2\geq a+b-\frac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

Ta có:

\(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)

\(\Rightarrow\)  \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)

Vì   \(a+b+c+d\ne0\)  nên   \(a=b=c=d\)

Do đó:   \(M=4\)

M =4 nha . TICK MÌNH ĐI !!!!!!!!!!!!!!!!!!!!!!

Bạn nhân chéo rồi PTNT là ok

a: ad=bc

=>a/b=c/d=k

=>a=bk; c=dk

b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)

a/b=bk/b=k

=>(a+c)/(b+d)=a/b

c: ad=bc

nên a/c=b/d

d: \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=k+1\)

\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=k+1\)

=>\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Rightarrow\left[\left(a+d\right)+\left(b+c\right)\right]\left[\left(a+d\right)-\left(b+c\right)\right]-\left[\left(a-d\right)-\left(b-c\right)\right]\left[\left(a-d\right)+\left(b-c\right)\right]=0\)

\(\Rightarrow\left(a+d\right)^2-\left(b+c\right)^2-\left(a-d\right)^2+\left(b-c\right)^2=0\)

\(\Rightarrow a^2+d^2+2ad-b^2-c^2-2bc-a^2-d^2+2ad+b^2+c^2-2bc\)

\(\Rightarrow4ad-4bc\)

\(\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\)