Với \(a,b>0\), chứng minh: \(a+\frac{b}{\left(a-b\right)\left(b+1\right)^2}\ge3\)
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\(VT=a+\frac{4}{\left(a-b\right)\left(b+1\right)^2}\)
\(=a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+\frac{b+1}{2}+\frac{b+1}{2}-1\)
\(\ge4\sqrt[4]{a-b\cdot\frac{4}{\left(a-b\right)\left(b+1\right)^2}\cdot\frac{b+1}{2}\cdot\frac{b+1}{2}}-1\)
\(\ge4-1=3=VP\)
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Câu hỏi của tran duc huy - Toán lớp 10 | Học trực tuyến
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Áp dụng BĐT AM-GM ta có:
\(VT=a^2+b^2+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b\)
\(=1+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b\)
\(=1+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{1}{a}+2a\right)+\left(\frac{1}{b}+2b\right)-\left(a+b\right)\)
\(\ge3+2\sqrt{\frac{1}{a}\cdot2a}+2\sqrt{\frac{1}{b}\cdot2b}-\sqrt{2\left(a^2+b^2\right)}\)
\(\ge3+4\sqrt{2}-\sqrt{2}=3+3\sqrt{2}=3\left(1+\sqrt{2}\right)\)
Khi \(a=b=\frac{1}{\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+\frac{4}{b\left(a-b\right)^2}=a-b+b+\frac{4}{b\left(a-b\right)^2}\ge a-b+2\sqrt{\frac{4b}{b\left(a-b\right)^2}}=a-b+\frac{4}{a-b}\ge4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=3\\b=1\end{matrix}\right.\)
b/ \(a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+b\ge2\sqrt{\frac{4\left(a-b\right)}{\left(a-b\right)\left(b+1\right)^2}}+b=\frac{4}{b+1}+b+1-1\ge4-1\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải
Áp dụng BĐT AM-GM:
\(\text{VT}=(a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}-1\geq 4\sqrt[4]{1}-1=3\)
Do đó ta có đpcm
Dấu $=$ xảy ra khi $b=1,a=2$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\Sigma_{cyc}a^3\left(b-c\right)}{\Sigma_{cyc}a^2\left(b-c\right)}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}=a+b+c\ge3\sqrt[3]{abc}\)