Question

Cho a,b > 0 và \(a+b\ge2\) Tìm Giá trị nhỏ nhất của :

\(M=\frac{a^3}{\left(b+1\right)^2}+\frac{b^3}{\left(a+1\right)^2}\)

Answer 1

M=\(\frac{a^4}{a\left(b+1\right)^2}+\frac{b^4}{b\left(a+1\right)^2}\)

áp dụng bdt bunhiacopxki ta co

(a+b)M>=\(\left(\frac{a^2}{b+1}+\frac{b^2}{a+1}\right)^2\)

\(\left(\frac{a^2}{b+1}+\frac{b^2}{a+1}\right)^2>=\left[\frac{\left(a+b^2\right)}{a+1+b+1}\right]^2\)

\(=\frac{\left(a+b\right)^4}{\left(a+b+2\right)^2}>=\frac{\left(a+b\right)^4}{4\left(a+b\right)^2}\)(do 2<=a+b)

=\(\frac{\left(a+b\right)^2}{4}\)

do do M(a+b)>=\(\frac{\left(a+b\right)^2}{4}\)

=>M>=\(\frac{a+b}{4}>=\frac{1}{2}\)

dau = xay ra <=> a=b=1