cho a,b,c > 0 chứng minh : [tex]\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}\geqslant \frac{a-d}{a+b}[/tex]
[tex]\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}\geqslant \frac{a-d}{a+b}\\\\ <=> \frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+ \frac{d-a}{a+b}\geq 0\\\\ <=> \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\geq 4\\\\ <=> (a+c).(\frac{1}{b+c}+\frac{1}{a+d})+(b+d).(\frac{1}{c+d}+\frac{1}{a+b})\geq 4[/tex] [tex]\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}\geqslant \frac{a-d}{a+b}\\\\ <=> \frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+ \frac{d-a}{a+b}\geq 0\\\\ <=> \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\geq 4\\\\ <=> (a+c).(\frac{1}{b+c}+\frac{1}{a+d})+(b+d).(\frac{1}{c+d}+\frac{1}{a+b})\geq 4[/tex]
mà : [tex](a+c).(\frac{1}{b+c}+\frac{1}{a+d})+(b+d).(\frac{1}{c+d}+\frac{1}{a+b})\\\\ \geq (a+c).\frac{4}{a+b+c+d}+(b+d).\frac{4}{a+b+c+d}\\\\ =\frac{4}{a+b+c+d}.(a+b+c+d)=4 (đpcm)[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
