Question

Cho a,b là các số thực dương và ab=1. Chứng minh:

1/ \(\frac{1}{a^3+5}+\frac{1}{b^3+5}\le\frac{1}{3}\)                                    2/ \(\frac{a^2}{b}+\frac{b^2}{a}+\frac{8}{a^2+b^2+6}\ge3\)

Giúp mk gấp nha gần thi rồi

Answer 1

\(1)\) \(\frac{1}{a^3+5}+\frac{1}{b^3+5}\le\frac{1}{3a+3}+\frac{1}{3b+3}=\frac{1}{3}\left(\frac{1}{a+1}+\frac{1}{b+1}\right)=\frac{1}{3}\left[\frac{a+b+2}{\left(a+1\right)\left(b+1\right)}\right]\)

\(=\frac{1}{3}\left(\frac{ab+a+b+1}{ab+a+b+1}-\frac{ab-1}{ab+a+b+1}\right)=\frac{1}{3}\left(1-\frac{0}{ab+a+b+1}\right)=\frac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}ab=1\\a^3=1\\b^3=1\end{cases}\Leftrightarrow a=b=1}\)

Answer 2

2) bđt \(\Leftrightarrow\)\(\left(a-b\right)^2\left(\frac{1}{a}+\frac{1}{b}\right)+\left(a+b-2\right)\left[\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\frac{7}{2}\right]\ge0\) (1) 

(1) đúng do \(a+b\ge2\sqrt{ab}=2\)\(\Leftrightarrow\)\(a+b-2\ge0\)

Dấu "=" xảy ra khi a=b=1