[Toán 7] Toán nâng cao

I

iceghost

Ta có : $d > 0$
$\implies a+b+c+d > a+b+c \\
\implies \dfrac{a}{a+b+c+d}<\dfrac{a}{a+b+c}$
Tương tự : $\dfrac{b}{a+b+c+d}<\dfrac{b}{b+c+d} \\
\dfrac{c}{a+b+c+d}<\dfrac{c}{c+d+a} \\
\dfrac{d}{a+b+c+d}<\dfrac{d}{d+a+b}$
$\implies \dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d} = 1 < \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b} \; (1)$

Lại có : $b>0$
$\implies a+b+c > a+c \\
\implies \dfrac{a}{a+b+c} < \dfrac{a}{a+c}$
Tương tự : $\dfrac{b}{b+c+d} < \dfrac{b}{b+d} \\
\dfrac{c}{c+d+a} < \dfrac{c}{a+c} \\
\dfrac{d}{d+a+b} < \dfrac{d}{b+d} \\
\implies \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+ \dfrac{d}{d+a+b} < \dfrac{a}{a+c}+\dfrac{b}{b+d}+\dfrac{c}{a+c}+ \dfrac{d}{b+d}=2 \; (2)$

Từ $(1), (2) \implies đpcm$
 
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