Ta có:
`P=a+b+1/a+1/b`
`=16/16a+25/25b+1/a+1/b`
`=(1/16a+1/a)+(1/25b+1/b)+15/16a+24/25b`
\(\left\{{}\begin{matrix}a\ge4\\b\ge5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{16}a>0;\dfrac{1}{a}>0\\\dfrac{1}{25}b>0;\dfrac{1}{b}>0\end{matrix}\right.\)
Áp dụng bđt cô-si ta có:
\(P\ge2\sqrt{\dfrac{1}{16}a\cdot\dfrac{1}{a}}+2\sqrt{\dfrac{1}{25}b\cdot\dfrac{1}{b}}+\dfrac{15}{16}a+\dfrac{24}{25}b\\ \ge2\cdot\dfrac{1}{4}+2\cdot\dfrac{1}{5}+\dfrac{15}{16}\cdot4+\dfrac{24}{25}\cdot5=\dfrac{189}{20}\)
Dấu "=" xảy ra khi: `a=4` và `b=5`
Đúng 1
Bình luận (0)
