Question

[Phạm Mỹ Châu]

cho a,b,c,d >0 thoả mãn a+b+c+d = 4

cmr \(\frac{a}{1+b^2}\)+ \(\frac{b}{1+c^2}\)+\(\frac{c}{1+d^2}\)+ \(\frac{d}{1+a^2}\)\(\ge\)2

Answer 1

[Thiên An]

Ta có  \(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự  \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\)

\(\frac{c}{1+d^2}\ge c-\frac{cd}{2}\)

\(\frac{d}{1+a^2}\ge d-\frac{ad}{2}\)

Lại có  \(ab+bc+cd+da\le\frac{\left(a+b+c+d\right)^2}{4}=\frac{4^2}{4}=4\)

Do đó  \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2}\ge\left(a+b+c+d\right)-\frac{ab+bc+cd+da}{2}\)

\(\ge4-\frac{4}{2}=2\)

Đẳng thức xảy ra  \(\Leftrightarrow\)  \(a=b=c=d=1\)