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

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

Theo đề, ta có:

\(\left\{{}\begin{matrix}a\ge c+d\\b\ge c+d\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-c\ge d\ge0\\b-d\ge c\ge0\end{matrix}\right.\) 

\(\Rightarrow\left(a-c\right)\left(b-d\right)\ge cd\)

\(\Leftrightarrow ab-bc-ad+cd\ge cd\)

\(\Leftrightarrow\) \(ab\ge ad+bc\left(đpcm\right)\)

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

\(\Rightarrow ad< bc\)

\(\Rightarrow ad+ab< bc+ab\)

\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

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

Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\)

ca này để thầy lâm ròi:<

\(\left(a^2+b+c+d\right)\left(1+b+c+d\right)\ge\left(a+b+c+d\right)^2=16\)

\(\Rightarrow\dfrac{1}{a^2+b+c+d}\le\dfrac{1+b+c+d}{16}=\dfrac{5-a}{16}\)

Tương tự: \(\dfrac{1}{b^2+c+d+a}\le\dfrac{5-b}{16}\) ...

Cộng vế:

\(P\le\dfrac{20-\left(a+b+c+d\right)}{16}=1\)

Dấu "=" xảy ra khi \(a=b=c=d=1\)

Để  ( a ; b ) ⊂ ( c ; d ) thì  c ≤ a < b ≤ d

Đáp án D