Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a-b+b+\frac{1}{b\left(a-b\right)}\ge3\sqrt[3]{\frac{\left(a-b\right)b.1}{b\left(a-b\right)}}=3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(VT=a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+\frac{b+1}{2}+\frac{b+1}{2}-1\)
\(VT\ge4\sqrt[4]{\frac{4\left(a-b\right)\left(b+1\right)^2}{4\left(a-b\right)\left(b+1\right)^2}}-1=3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=1\\a=2\end{matrix}\right.\)
\(\frac{a-b}{2}+\frac{a-b}{2}+\frac{1}{b\left(a-b\right)^2}+b\ge4\sqrt[4]{\frac{b\left(a-b\right)^2}{4b\left(a-b\right)^2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=\frac{3\sqrt{2}}{2}\\b=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
Bạn tham khảo:
Câu hỏi của tran duc huy - Toán lớp 10 | Học trực tuyến
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
\(a+\frac{4}{b\left(a-b\right)^2}=a-b+b+\frac{4}{b\left(a-b\right)^2}\ge a-b+2\sqrt{\frac{4b}{b\left(a-b\right)^2}}=a-b+\frac{4}{a-b}\ge4\)
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 \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm
1. Ta có: \(a-b+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\ge\dfrac{4}{b+1}\)
\(a+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\ge\dfrac{4}{b+1}+b\)(1)
lại có: \(\dfrac{4}{b+1}+b+1\ge4\)
\(\dfrac{4}{b+1}+b\ge3\)(2)
Từ (1),(2) ta có:\(a+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\ge3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-b=\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\\b+1=\dfrac{4}{b+1}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
2. Ta có\(\dfrac{2a^3+1}{4b\left(a-b\right)}\ge3\)
\(\Leftrightarrow2a^3+1\ge12ab-12b^2\)
\(\Leftrightarrow2a^3+1-12ab+12b^2\ge0\)
\(\Leftrightarrow2a^3-3a^2+1+3\left(a-2b\right)^2\ge0\)
\(\Leftrightarrow\left(2a+1\right)\left(a-1\right)^2+3\left(a-2b\right)^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-1=0\\a-2b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=\dfrac{1}{2}\end{matrix}\right.\)
cảm ơn bạn