Cho a, b là các số dương thỏa mãn a + b = 3. Chứng minh rằng:

Xét vế trái

${\left(a+\frac{1}{b}\right)}^2+{\left(b+\frac{1}{a}\right)}^2$

$$\begin{gathered} =a^2+2\frac{a}{b}+\frac{1}{b^2}+b^2+2\frac{b}{a}+\frac{1}{a^2} \\ =\left(\frac{16}{81}{a}^2+\frac{1}{a^2}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{16}{81}{b}^2+\frac{1}{b^2}\right)+\frac{65}{81}(a^2+b^2) \end{gathered}$$

$$\begin{gathered} =\left(\frac{16}{81}{a}^2+\frac{1}{a^2}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{16}{81}{b}^2+\frac{1}{b^2}\right)+\frac{65}{81}\left[a^2+{\left(3-a\right)}^2\right] \\ =\left(\frac{16}{81}{a}^2+\frac{1}{a^2}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{16}{81}{b}^2+\frac{1}{b^2}\right)+\frac{65}{81}\left[2a^2-6a+9\right] \end{gathered}$$

$$\begin{gathered} =\left(\frac{16}{81}{a}^2+\frac{1}{a^2}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{16}{81}{b}^2+\frac{1}{b^2}\right)+\frac{65}{81}\left[2a^2-6a+9\right] \\ =\left(\frac{16}{81}{a}^2+\frac{1}{a^2}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{16}{81}{b}^2+\frac{1}{b^2}\right)+\frac{130}{81}\left[a^2-2.\frac{3}{2}a+\frac{9}{4}+\frac{9}{4}\right] \end{gathered}$$

$$\begin{gathered} =\left(\frac{16}{81}{a}^2+\frac{1}{a^2}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{16}{81}{b}^2+\frac{1}{b^2}\right)+\frac{130}{81}{\left(a-\frac{3}{2}\right)}^2+\frac{65}{18} \\ {\ge }^{\cos i}2\sqrt{\frac{16}{81}\frac{a^2}{a^2}}+2.2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{16}{81}\frac{b^2}{b^2}}+0+\frac{65}{18} \end{gathered}$$

=$\frac{169}{18}$ (điều phải chứng minh).

Dấu “=” xảy ra khi a = b = $\frac{3}{2}$ .