Question

cho a,b,c,d là các số nguyên dương. CMR

1<a/(a+b+c)+b/(b+c+d)+c/(c+d+a)+d/(d+a+b)<2

Answer 1

Vì a,b,c,d thuộc N*

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

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

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

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

e cộng vế theo vế đc 1<...<2

Answer 2

Ta có \(\frac{a}{a+b+c}>\frac{a}{a+b+c+d} \quad (vì\quad a,b,c,d>0)\)

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

=> \(\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}>\frac{a+b+c+d}{a+b+c+d}=1\) (1)

Lại có:\(\frac{a}{a+b+c}<\frac{a}{a+b} \quad (vì\quad a,b,c,d>0)\);

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

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

Từ (1) và (2) Ta có...

Answer 3

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

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

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

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

Cộng vế theo vế ta được \(\frac{a}{a+b+c}\)+\(\frac{b}{d+b+c}\)+\(\frac{c}{a+d+c}\)+\(\frac{d}{a+b+d}\)>1    (1)

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

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

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

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

Cộng vế theo vế ta được \(\frac{a}{a+b+c}\)+\(\frac{b}{d+b+c}\)+\(\frac{c}{c+d+a}\)+\(\frac{d}{d+a+b}\)<2    (2)

Từ (1) và (2) suy ra ghi lại đề bài (đpcm )