\[P = \frac{{{x^3} + {y^2}}}{{{x^2}}} + \frac{{{x^2} + {y^3}}}{{{y^2}}} + \frac{3}{{2x}} + \frac{3}{{2y}}\]\[ = x + \frac{{{y^2}}}{{{x^2}}} + \frac{{{x^2}}}{{{y^2}}} + y + \frac{3}{2}(\frac{1}{x} + \frac{1}{y})\]
Ta có:
\[\frac{{{y^2}}}{{{x^2}}} + \frac{{{x^2}}}{{{y^2}}} \ge 2\sqrt {\frac{{{y^2}}}{{{x^2}}}.\frac{{{x^2}}}{{{y^2}}}} = 2\]
\[\frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}} = \frac{4}{2} = 2\]
$ \Rightarrow P \ge 2 + 2 + \frac{3}{2}.2 = 7$. Đẳng thức xảy ra $ \Leftrightarrow x = y = 1$