Since $x,y \in \left[1,\frac{3}{2} \right]$ the $ 3-2x $ and $3-2y$ are positive. So, we can apply A.M.-G.M. inequality, to get $$ \frac{x^2 + (3-2y)}{2} \enspace \quad \geq \quad \enspace x\sqrt{3-2y}$$ and $$ \frac{y^2 +(3-2x)}{2} \enspace \quad \geq \quad \enspace y\sqrt{3-2x}$$ Adding above two inequalities, we have $$\frac{x^2+3-2y+y^2+3-2x}{2} \quad \enspace \geq \quad \enspace y\sqrt{3-2x}+x\sqrt{3-2y}$$ Therefore, it suffices to show that $$x^2 \enspace + \enspace y^2 \quad \geq \quad \frac{x^2+3-2y+y^2+3-2x}{2}$$ $$ \iff \quad \quad x^2 \enspace+ \enspace 2x \enspace - \enspace 3 \enspace + \enspace y^2 \enspace + \enspace 2y \enspace - \enspace 3 \quad \geq \quad 0$$ $$ \iff \quad \quad (x-1)(x+3) \enspace + \enspace (y-1)(y+3) \enspace \geq \enspace 0$$ which is true since $x,y \geq 1$. Hence, proved.