Đặt [tex]a=tb[/tex]
Ta có
[tex]t\sqrt{t}b^2-\sqrt{t}b^2=tb+b\\\Leftrightarrow b(t\sqrt{t}-\sqrt{t})=t+1\\\Leftrightarrow b=\frac{t+1}{\sqrt{t}(t-1)}\\\Rightarrow a=\frac{t^2+t}{\sqrt{t}(t-1)}\\\Rightarrow a+b=\frac{(t-1)^2}{\sqrt{t}(t-1)}+\frac{4t}{\sqrt{t}(t-1)}\geq 4[/tex]
Đặt [tex]a=tb[/tex]
Ta có
[tex]t\sqrt{t}b^2-\sqrt{t}b^2=tb+b\\\Leftrightarrow b(t\sqrt{t}-\sqrt{t})=t+1\\\Leftrightarrow b=\frac{t+1}{\sqrt{t}(t-1)}\\\Rightarrow a=\frac{t^2+t}{\sqrt{t}(t-1)}\\\Rightarrow a+b=\frac{(t-1)^2}{\sqrt{t}(t-1)}+\frac{4t}{\sqrt{t}(t-1)}\geq 4[/tex]