Question

CM:

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{a+d}+\dfrac{d}{a+b}\ge2\)

Biết a; b; c; d >0

Answer 1

Lời giải:

Áp dụng BĐT Cauchy- Schwarz:

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

Lại có:

\((a+b+c+d)^2=[(a+c)+(b+d)]^2=(a+c)^2+(b+d)^2+2(a+c)(b+d)\)

Áp dụng BĐT Am-Gm:

\((a+c)^2+(b+d)^2\geq 4ac+4bd\)

\(\Rightarrow (a+b+c+d)^2\geq 4ac+4bd+2(ab+bc+cd+da)\)

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

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c=d>0\)