Question

Cho a,b,c,d >0 .CMR: a/(b+c) + b/(c+d) + c/(d+a) + d/( a+b)>=2

Answer 1

Áp dụng BĐT Bunhiacopxki , ta có: 

Với a,b,c,d >0

\(\left(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\right)\left[a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)\right]\ge\left(a+b+c+d\right)^2\)

\(\Rightarrow\left(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\right)\ge\frac{\left(a+b+c+d\right)^2}{ab+bc+cd+da+2ca+2bd}\)

Ta cần chứng minh : 

\(\left(a+b+c+d\right)^2\ge2\left(ab+bc+cd+da+2ac+2bd\right)\)

\(\Leftrightarrow a^2+b^2+c^2+d^2\ge2ca+2bd\)

\(\Leftrightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\)(đúng) 

\(\Leftrightarrow dpcm\)