Question

 

 

1,Chứng minh:

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

2, Tìm min của D=\(\sqrt{x^2+1}+\sqrt{x^2-2x+5}\)

Answer 1

1,

\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+bd}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{\left(a+c\right)\left(b+d\right)+2ac+2bd}=\frac{2\left(a+c\right)\left(b+d\right)+\left(a+c\right)^2+\left(b+d\right)^2}{\left(a+c\right)\left(b+d\right)+2ac+2bd}\)

\(\ge\frac{2\left(a+c\right)\left(b+d\right)+4ac+4bd}{\left(a+c\right)\left(b+d\right)+2ac+2bd}=2\)

Answer 2

\(\sqrt{x^2+1}+\sqrt{\left(1-x\right)^2+2^2}\ge\sqrt{\left(x+1-x\right)^2+\left(1+2\right)^2}=\sqrt{10} .\\ \)
Dấu ''='' \(x=\frac{1}{3}\\ \)
BĐT phụ: \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\\ \)
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