@Akai Haruma

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#)Giải :

Ta có : \(c\left(b+d\right)=2bd\Rightarrow bc+cd=2bd\Rightarrow\frac{bc+cd}{a+c}=\frac{2bd}{2b}=d\)

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\hept{\begin{cases}\left(\frac{a}{b}\right)^8=\left(\frac{c}{d}\right)^8=\left(\frac{a+c}{b+d}\right)^8=\frac{a^8}{b^8}\\\left(\frac{a}{b}\right)^8=\left(\frac{c}{d}\right)^8=\frac{a^8}{b^8}=\frac{c^8}{d^8}=\frac{a^8+c^8}{b^8+c^8}\end{cases}}\)

\(\Rightarrow\left(\frac{a+c}{b+d}\right)^8=\frac{a^8+b^8}{c^8+d^8}\left(đpcm\right)\)

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

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

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

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

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

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

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

Lại có: \(\frac{a}{a+b+c}< \frac{a+c}{a+b+c+d}\)

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

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

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

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

\(\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{2a+2b+2c+2d}{a+b+c+d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

\(\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2\)        (2)

Từ (1)(2) => \(1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2\)   (đpcm)