ĐK:[tex]-\sqrt{2}\leq x\leq \sqrt{2}[/tex]
Ta thấy:[tex]x+\sqrt{2-x^2}\leq \sqrt{2(x^2+2-x^2)}=\sqrt{4}=2[/tex]
[tex]4y^2+4y+3=(2y+1)^2+2\geq 2[/tex] [tex]\Rightarrow VT\leq 2\leq VP[/tex]
Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} x=\sqrt{2-x^2}\\ 2y+1=0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=-\frac{1}{2} \end{matrix}\right.[/tex]