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) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)

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: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)

Bạn tham khảo:

Câu hỏi của tran duc huy - Toán lớp 10 | Học trực tuyến

Áp dụng BĐT Cô-si cho 3 số dương ta có:

\(\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(\sqrt[3]{\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)}\right)^4\)

Ta chứng minh: \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\ge\left(1+\frac{3}{2+abc}\right)^3\left(1\right)\)

Theo BĐT Cô - si ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\)

\(\ge1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{\left(abc\right)^2}}+\frac{1}{abc}=\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\ge\left(1+\frac{3}{2+abc}\right)^3\)

(Vì \(abc+2=abc+1+1\ge3\sqrt[3]{abc}\))

Vậy \(\left(1\right)\) được chứng minh \(\Rightarrow BĐT\) đúng \(\forall a,b,c>0\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=1\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow VT\ge3\sqrt[3]{\left[\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\right]^4}\)

\(\Rightarrow VT\ge3\left(\sqrt[3]{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}}\right)^4\left(1\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\\\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3\sqrt[3]{\frac{1}{a^2b^2c^2}}\end{cases}}\)

\(\Rightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge1+3\sqrt[3]{\frac{1}{abc}}\)

\(+3\sqrt[3]{\frac{1}{a^2b^2c^2}}+\frac{1}{abc}\)

\(\Rightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\)

\(\Rightarrow3\left(\sqrt[3]{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}}\right)^4\)

\(\ge3\left(1+\frac{1}{\sqrt[3]{abc}}\right)^4\)

\(\left(2\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\sqrt[3]{abc}\le\frac{abc+1+1}{3}=\frac{abc+2}{3}\)

\(\Rightarrow1+\frac{1}{\sqrt[3]{abc}}\ge1+\frac{3}{abc+2}\)

\(\Rightarrow3\left(1+\frac{1}{\sqrt[3]{abc}}\right)^4\ge3\left(1+\frac{3}{abc+2}\right)^4\left(3\right)\)

Từ (1) , (2) và (3) 

\(\Rightarrow VT\ge3\left(1+\frac{3}{abc+2}\right)^4\)

\(\Leftrightarrow\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(1+\frac{3}{2+abc}\right)^4\left(đpcm\right)\)

Chúc bạn học tốt !!!

Vì nó thik thì nó \(\ge\) thôi

Đúng 100%

Đúng 100%

Đúng 100%

\(a+\frac{4}{b\left(a-b\right)^2}=a-b+b+\frac{4}{b\left(a-b\right)^2}\ge a-b+2\sqrt{\frac{4b}{b\left(a-b\right)^2}}=a-b+\frac{4}{a-b}\ge4\)

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 hay quá!

Theo bất đẳng thức Cô-Si cho 3 số dương ta có

\(\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\sqrt[3]{\left(1+\frac{1}{a}\right)^4\left(1+\frac{1}{b}\right)^4\left(1+\frac{1}{c}\right)^4}\).

Do đó ta chỉ cần chứng minh \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\ge\left(1+\frac{3}{2+abc}\right)^3.\) (Lúc đó kết hợp hai bất đẳng thức ta được ngay điều phải chứng minh).

Thực vậy, đầu tiên áp dụng bất đẳng thức Cô-Si cho 3 số dương ta có

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=1+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{1}{abc}\ge\)

\(\ge1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{a^2b^2c^2}}+\frac{1}{abc}=\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3.\)

Mặt khác ta có \(2+abc=1+1+abc\ge3\sqrt[3]{abc}\to\frac{1}{\sqrt[3]{abc}}\ge\frac{3}{2+abc}\to\)

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\ge\left(1+\frac{3}{2+abc}\right)^3.\)    (ĐPCM)

WTF Toán Lớp 1

thấy mẹ nhầm rồi,  quy đồng quên nhân:(( mai rảnh check lại:((

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 \(a=b=\frac{1}{\sqrt{2}}\)