\(VT=\dfrac{\left(a+c\right)^2}{\left(a+c\right)\left(a+b\right)}+\dfrac{\left(b+d\right)^2}{\left(b+c\right)\left(b+d\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(c+d\right)}+\dfrac{\left(d+b\right)^2}{\left(d+a\right)\left(d+b\right)}\)

\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)

Dấu "=" xảy ra khi \(a=b=c=d\)

Vì \(0\le a\le b\le c\le1\) nên:

\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)

Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)

Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)

Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)

Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)

\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\\ \Rightarrow ad+ab< bc+ab\\ \Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

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

Theo tính chất dãy tỉ số bằng nhau, đặt:

\(\dfrac{a}{A}=\dfrac{b}{B}=\dfrac{c}{C}=\dfrac{d}{D}=\dfrac{a+b+c+d}{A+B+C+D}=k>0\)

\(\Rightarrow a=kA;b=kB;c=kC;d=kD;a+b+c+d=k\left(A+B+C+D\right)\)

Do đó:

\(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{kA^2}+\sqrt{kB^2}+\sqrt{kC^2}+\sqrt{kD^2}\)

\(=\sqrt{k}\left(A+B+C+D\right)\) (1)

\(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{k\left(A+B+C+D\right)^2}=\sqrt{k}\left(A+B+C+D\right)\) (2)

Từ (1);(2) suy ra điều phải c/m

\(\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}\le1-\dfrac{a}{1+a}=\dfrac{1}{1+a}\)

\(\Rightarrow\dfrac{1}{1+a}\ge\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}\ge3\dfrac{\sqrt[3]{bcd}}{\sqrt[3]{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)

Chứng minh tương tự ta có:

\(\dfrac{1}{1+b}\ge3\dfrac{\sqrt[3]{acd}}{\sqrt[3]{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+c}\ge3\dfrac{\sqrt[3]{abd}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+d}\ge3\dfrac{\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Nhân vế với vế của các BĐT trên ta được:

\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\dfrac{abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)

\(\Rightarrow81abcd\le1\Rightarrow abcd\le\dfrac{1}{81}\)

Dấu "=" xảy ra khi \(a=b=c=d=\dfrac{1}{3}\)

Không mất tính tổng quát, giả sử \(2\ge a\ge b\ge c\ge1\)

Khi đó dễ thấy dấu = sẽ đạt được tại biên, tức a=2, c=1 nên ta sẽ dồn các biến ra biên

Ta có: \(\left(\dfrac{a}{b}-1\right)\left(\dfrac{b}{c}-1\right)\ge0\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{c}\le\dfrac{a}{c}+1\)

\(\left(\dfrac{b}{a}-1\right)\left(\dfrac{c}{b}-1\right)\ge0\Leftrightarrow\dfrac{b}{a}+\dfrac{c}{b}\le\dfrac{c}{a}+1\)

Do đó \(VT\le2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+2\) nên chỉ cần chứng minh \(\dfrac{a}{c}+\dfrac{c}{a}\le\dfrac{5}{2}\)(*) hay \(\dfrac{\left(a-2c\right)\left(2a-c\right)}{2ac}\le0\) ( luôn đúng do \(c\le a\le2c\) )

Vậy ta có đpcm. Dấu = xảy ra khi a=2, c=1, b=1 hoặc a=2, c=1, b=2 và các hoán vị tương ứng.

Lời giải:
a. 

$\frac{a}{b}< \frac{c}{d}\Rightarrow \frac{a}{b}-\frac{c}{d}<0$

$\Rightarrow \frac{ad-bc}{bd}< 0$

$\Rightarrow ad-bc<0$ (do $bd>0$)

$\Rightarrow ad< bc$ (đpcm)

b.

$\frac{a}{b}-\frac{a+c}{b+d}=\frac{a(b+d)-b(a+c)}{b(b+d)}=\frac{ad-bc}{b(b+d)}<0$ do $ad-bc<0$ và $b(b+d)>0$

$\Rightarrow \frac{a}{b}< \frac{a+c}{b+d}$

--------

$\frac{a+c}{b+d}-\frac{c}{d}=\frac{d(a+c)-c(b+d)}{d(b+d)}=\frac{ad-bc}{d(b+d)}<0$ do $ad-bc<0$ và $d(b+d)>0$

$\Rightarrow \frac{a+c}{b+d}< \frac{c}{d}$
Ta có đpcm.