Câu hỏi của Hồ Đình Bảo Long

Question

cho a,b,c,d >0

cmr: a-d/d+b+d-b/b+c+b-c/c+a+c-a/a+d>=0

Đinh Đức Hùng · Accepted Answer

Cộng 4 vào vế trái nhá

$VT+4=\left(\dfrac{a-d}{d+b}+1\right)+\left(\dfrac{d-b}{b+c}+1\right)+\left(\dfrac{b-c}{c+a}+1\right)+\left(\dfrac{c-a}{a+d}+1\right)$

$=\dfrac{a+b}{d+b}+\dfrac{d+c}{b+c}+\dfrac{a+b}{c+a}+\dfrac{c+d}{a+d}$

$=\left(a+b\right)\left(\dfrac{1}{d+b}+\dfrac{1}{c+a}\right)+\left(c+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)$

$\ge\left(a+b\right).\dfrac{4}{a+b+c+d}+\left(c+d\right).\dfrac{4}{a+b+c+d}$

$=\left(a+b+c+d\right).\dfrac{4}{a+b+c+d}$$=4$

$\Rightarrow VT\ge0=VP$(Đpcm)Câu trả lời: Cộng 4 vào vế trái nhá

$VT+4=\left(\dfrac{a-d}{d+b}+1\right)+\left(\dfrac{d-b}{b+c}+1\right)+\left(\dfrac{b-c}{c+a}+1\right)+\left(\dfrac{c-a}{a+d}+1\right)$

$=\dfrac{a+b}{d+b}+\dfrac{d+c}{b+c}+\dfrac{a+b}{c+a}+\dfrac{c+d}{a+d}$

$=\left(a+b\right)\left(\dfrac{1}{d+b}+\dfrac{1}{c+a}\right)+\left(c+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)$

$\ge\left(a+b\right).\dfrac{4}{a+b+c+d}+\left(c+d\right).\dfrac{4}{a+b+c+d}$

$=\left(a+b+c+d\right).\dfrac{4}{a+b+c+d}$$=4$

$\Rightarrow VT\ge0=VP$(Đpcm)