Bài 1 : Cho \(a>b>0\)
CMR : \(a+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\ge3\)
Dấu "=" xảy ra khi nào
Bài 2 : Cho \(a,b>0\)
CM : \(\dfrac{2a^3+1}{4b\left(a-b\right)}\ge3\)
Dấu "=" xảy ra khi nào
K
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NT
Nguyễn thành Đạt
CTVHS
20 tháng 3 2023
3.1
Xét hiệu :
\(\left(\dfrac{a+b}{2}\right)^2-ab=\dfrac{a^2+2ab+b^2}{4}-\dfrac{4ab}{4}\)
\(=\dfrac{a^2-2ab+b^2}{4}=\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\in R\)
Vậy \(\left(\dfrac{a+b}{2}\right)^2\ge ab,\forall a,b\in R\)
Dấu bằng xảy ra : \(\Leftrightarrow a=b\)
3.2
Áp dụng kết quả của câu 3.1 vào câu 3.2 ta được:
\(\left(a+b+c\right)^2=[a+\left(b+c\right)]^2\ge4a\left(b+c\right)\)
Mà : \(a+b+c=1\left(gt\right)\)
nên : \(1\ge4a\left(b+c\right)\)
\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\) ( vì a,b,c không âm nên b+c không âm )
Mà : \(\left(b+c\right)^2\ge4bc\Leftrightarrow\left(b-c\right)^2\ge0,\forall b,c\in N\)
\(\Rightarrow b+c\ge16abc\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c\end{matrix}\right.\Leftrightarrow b=c=\dfrac{1}{4};a=\dfrac{1}{2}\)
S
27 tháng 8 2021
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
27 tháng 8 2021
a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 12 2020
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Do \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}=\dfrac{2a^2}{ab+ac}+\dfrac{2b^2}{bc+ab}+\dfrac{2c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Điều này hiển nhiên đúng do:
\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)
\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)
Dấu "=" xảy ra khi \(a=b=c\)
KK
13 tháng 5 2017
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow VT\ge3\sqrt[3]{\left[\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\right]^4}\)
\(\Rightarrow VT\ge3\left(\sqrt[3]{1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}}\right)^4\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\\\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge3\sqrt[3]{\dfrac{1}{a^2b^2c^2}}\end{matrix}\right.\)
\(\Rightarrow1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}\ge1+3\sqrt[3]{\dfrac{1}{abc}}+3\sqrt[3]{\dfrac{1}{a^2b^2c^2}}+\dfrac{1}{abc}\)
\(\Rightarrow1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)
\(\Rightarrow3\left(\sqrt[3]{1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}}\right)^4\ge3\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^4\) (2)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt[3]{abc}\le\dfrac{abc+1+1}{3}=\dfrac{abc+2}{3}\)
\(\Rightarrow1+\dfrac{1}{\sqrt[3]{abc}}\ge1+\dfrac{3}{abc+2}\)
\(\Rightarrow3\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^4\ge3\left(1+\dfrac{3}{abc+2}\right)^4\) (3)
Từ (1) và (2) và (3)
\(\Rightarrow VT\ge3\left(1+\dfrac{3}{abc+2}\right)^4\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{abc+2}\right)^4\) ( đpcm )
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 6 2021
Bài này đã có ở đây:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
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 \(\left\{{}\begin{matrix}a-1=0\\a-2b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=\dfrac{1}{2}\end{matrix}\right.\)
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