24 tháng 5 2019 lúc 20:30
Đặt \(\left\{{}\begin{matrix}\sqrt{4a+1}=x\\\sqrt{4b+1}=y\end{matrix}\right.\) \(\Rightarrow1\le x;y\le3\)
\(\Rightarrow x^2+y^2=4\left(a+b\right)+2=10\)
Do \(1\le x\le3\Rightarrow\left(x-1\right)\left(x-3\right)\le0\Rightarrow x^2-4x+3\le0\)
\(\Rightarrow x^2+3\le4x\Rightarrow x\ge\frac{x^2+3}{4}\)
Tương tự, do \(1\le y\le3\Rightarrow y\ge\frac{y^2+3}{4}\)
\(\Rightarrow P=x+y\ge\frac{x^2+3}{4}+\frac{y^2+3}{4}=\frac{x^2+y^2+6}{4}=\frac{16}{4}=4\)
\(\Rightarrow P_{min}=4\) khi \(\left(x;y\right)=\left(1;3\right);\left(3;1\right)\) hay \(\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)
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