\(log_a\left(a^3b^2\right)=log_aa^3+log_ab^2=3+2\cdot log_ab\)

=>B

1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)

CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)

Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Thay vào biểu thwusc M ta được M=3abc (ĐPCM)

2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó

Nếu không thấy thì em có thể quy đồng lên mà rút gọn

vâng e cảm ơn anh 

Phát biểu C đúng.

Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)

\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)

Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

Ai có cách hay?

1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.

2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)

\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)

\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)

Thôi đành dồn về bậc dễ chịu hơn vậy :))
\(9=\frac{1}{a^3}+1+\frac{1}{a^3}+\frac{1}{b^3}+1+\frac{1}{b^3}+\frac{1}{c^3}+1+\frac{1}{c^3}\)

\(\ge\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le3\)

Đến đây ta có đánh giá bằng 2 cách như sau:

Cách 1:

Theo Bunhiacopski ta dễ có:

\(\left[2a+\left(b+c\right)\right]^2\ge4\cdot2a\left(b+c\right)\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{8a\left(b+c\right)}\)

\(\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{\left(b+c\right)^2}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{4bc}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8}\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\right]\)

Khi đó:

\(P\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8b^2}+\frac{1}{8c^2}+\frac{1}{4b^2}+\frac{1}{8a^2}+\frac{1}{8c^2}+\frac{1}{4c^2}+\frac{1}{8a^2}+\frac{1}{8b^2}\right]=\frac{3}{16}\)

Cách 2:

Áp dụng liên tiếp BĐT phụ dạng \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta dễ có rằng:

\(\frac{1}{\left(2a+b+c\right)^2}=\left(\frac{1}{2a+b+c}\right)^2=\frac{1}{16}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2=\frac{1}{16}\left[\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(a+c\right)^2}+\frac{2}{\left(a+b\right)\left(a+c\right)}\right]\)

\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(c+a\right)^2}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(b+c\right)\left(c+a\right)}+\frac{2}{\left(c+a\right)\left(a+b\right)}\)

\(\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)

\(\le4\cdot\frac{1}{16}\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)

\(=\frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\le\frac{1}{2}\cdot\left(3+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le3\)

\(\Rightarrow P\le\frac{3}{16}\)

Đẳng thức xảy ra tại a=b=c=1