Question

Bài 1: Cho \(a,b,c>0;a+b+c=3\)

Tìm min \(A=\dfrac{a^2}{b+1}+\dfrac{b^2}{c+1}+\dfrac{c^2}{a+1}\)

Bài 2: Cho \(a,b>0;a+b\le0\)

Tìm min \(B=ab+\dfrac{1}{ab}\)

Answer 1

\(A+\dfrac{1}{4}\left(a+b+c\right)+\dfrac{3}{4}=\left(\dfrac{a^2}{b+1}+\dfrac{1}{4}\left(b+1\right)\right)+\left(\dfrac{b^2}{c+1}+\dfrac{1}{4}\left(c+1\right)\right)+\left(\dfrac{c^2}{a+1}+\left(a+1\right)\right)\)\(A+\dfrac{3}{2}\ge a+b+c=3\Rightarrow A\ge\dfrac{3}{2}\)

Dấu "=" xảy ra <=> a = b = c = 1

Answer 2

a;b>0 and a+b<=0 ????

Answer 3

Mn ơi, ở câu 2 mk viết nhầm đề nhé. đề đúng là \(a+b\le1\) chứ không phải là\(a+b\le0\) nhé

Answer 4

\(B=ab+\dfrac{1}{ab}=\left(16ab+\dfrac{1}{ab}\right)-15ab\ge8-\dfrac{15}{4}.\left(a+b\right)^2\ge8-\dfrac{15}{4}.1=\dfrac{17}{4}\)

Dấu "=" xảy ra <=> \(a=b=\dfrac{1}{2}\)