Ta có:$(a-\frac{1}{2})^{2}=a^{2}-a+\frac{3}{4}-(a+b+\frac{1}{2})\Rightarrow a^{2}+b+\frac{3}{4}\geq a+b+\frac{1}{2}$TT$\Rightarrow VT\geq (a+b+\frac{1}{2})^{2}=(a+\frac{1}{4}+b+\frac{1}{4})^{2}\geq4(a+\frac{1}{4})(b+\frac{1}{4})=VP$Dấu''='' xra$\Leftrightarrow a=b=\frac{1}{2}$
Ta có:$(a-\frac{1}{2})^{2}=a^{2}
+b+\frac{3}{4}-(a+b+\frac{1}{2})\Rightarrow a^{2}+b+\frac{3}{4}\geq a+b+\frac{1}{2}$TT$\Rightarrow VT\geq (a+b+\frac{1}{2})^{2}=(a+\frac{1}{4}+b+\frac{1}{4})^{2}\geq4(a+\frac{1}{4})(b+\frac{1}{4})=VP$Dấu''='' xra$\Leftrightarrow a=b=\frac{1}{2}$