firmezas1

2022-09-06

1 College math competitions ranking

Let $a,b,c,d$ be real numbers which are strictly positive.

Show that

$1<{\displaystyle \frac{a}{a+b+c}}+{\displaystyle \frac{b}{b+a+d}}+{\displaystyle \frac{c}{c+a+d}}+{\displaystyle \frac{d}{d+c+b}}<2.$

I need help please.

Let $a,b,c,d$ be real numbers which are strictly positive.

Show that

$1<{\displaystyle \frac{a}{a+b+c}}+{\displaystyle \frac{b}{b+a+d}}+{\displaystyle \frac{c}{c+a+d}}+{\displaystyle \frac{d}{d+c+b}}<2.$

I need help please.

Mario Monroe

Beginner2022-09-07Added 12 answers

No advanced techniques are necessary, just observe that

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

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

Dangelo Rosario

Beginner2022-09-08Added 3 answers

Because $1=\sum _{cyc}\frac{a}{a+b+c+d}<\sum _{cyc}\frac{a}{a+b+c}<\sum _{cyc}\frac{a+d}{a+b+c+d}=2$

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