Đặt \(am^3=bn^3=cp^3=k\)

Ta có \(\sqrt[3]{k}=\sqrt[3]{a}m=\sqrt[3]{b}n=\sqrt[3]{c}p=\frac{\sqrt[3]{a}}{\frac{1}{m}}=\frac{\sqrt[3]{b}}{\frac{1}{n}}=\frac{\sqrt[3]{c}}{\frac{1}{p}}\)

\(=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\frac{1}{m}+\frac{1}{n}+\frac{1}{p}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\) \(\left(TCDTSBN\right)\)\(\left(1\right)\)

Ta cũng có \(k=\frac{am^2}{\frac{1}{m}}=\frac{bn^2}{\frac{1}{n}}=\frac{cp^2}{\frac{1}{p}}=\frac{am^2+bn^2+cp^2}{\frac{1}{m}+\frac{1}{n}+\frac{1}{p}}=am^2+bn^2+cp^2\)  \(\left(TCDTSBN\right)\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{k}\)

cách khác nhé: 

Đặt:   \(am^3=bn^3=cp^3=k^3\)

\(\Rightarrow\)\(a=\frac{k^3}{m^3};\)\(b=\frac{k^3}{n^3};\)\(c=\frac{k^3}{p^3}\)

Ta có:

\(VT=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

\(=\sqrt[3]{\frac{k^3}{m^3}}+\sqrt[3]{\frac{k^3}{n^3}}+\sqrt[3]{\frac{k^3}{p^3}}\)

\(=\frac{k}{m}+\frac{k}{n}+\frac{k}{p}=k\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)=k\)    (do 1/m + 1/n + 1/p = 1)

\(VP=\sqrt[3]{am^2+bn^2+cp^2}\)

\(=\sqrt[3]{\frac{k^3}{m^3}.m^2+\frac{k^3}{n^3}.n^2+\frac{k^3}{p^3}.p^2}\)

\(=\sqrt[3]{k^3\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)}=\sqrt[3]{k^3}=k\)   (do 1/m + 1/n + 1/p = 1)

suy ra:   \(VT=VP=k\) (đpcm)

=>\(am^3=bn^3=cp^3=\frac{am^3}{m}+\frac{bn^3}{n}+\frac{cp^3}{p}\)

=>\(am^3=bn^3=cp^3=am^2+bn^2+cp^2\)

\(\sqrt[3]{am^2+bn^2+cp^2}=m\sqrt[3]{a}=n\sqrt[3]{b}=p\sqrt[3]{c}\)

=>\(\sqrt[3]{am^2+bn^2+cp^2}.1=m\sqrt[3]{a}.\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)=\frac{m\sqrt[3]{a}}{m}+\frac{n\sqrt[3]{b}}{n}+\frac{p\sqrt[3]{c}}{p}\)

\(\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

chtt đi. tớ làm bài tương tự r 

chtt là cái j v?

\(A=\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{\frac{am^3}{m}+\frac{bn^3}{n}+\frac{cp^3}{p}}=\sqrt[3]{am^3\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)}\)
\(\Rightarrow A=\sqrt[3]{am^3}=m\sqrt[3]{a}\Rightarrow\frac{A}{m}=\sqrt[3]{a}\)
Tương tự :.. \(\Rightarrow A\left(\frac{1}{m}+\frac{1}{p}+\frac{1}{n}\right)=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

Đặt   \(am^3=bn^3=cp^3=k^3\)

\(\Rightarrow\)\(a=\frac{k^3}{m^3};\) \(b=\frac{k^3}{n^3};\) \(c=\frac{k^3}{p^3}\)

Ta có:  \(VT=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

                  \(=\sqrt[3]{\frac{k^3}{m^3}}+\sqrt[3]{\frac{k^3}{n^3}}+\sqrt[3]{\frac{k^3}{p^3}}\)

                  \(=\frac{k}{m}+\frac{k}{n}+\frac{k}{p}=k\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)=k\)

          \(VP=\sqrt[3]{am^2+bn^2+cp^2}\)

                 \(=\sqrt[3]{\frac{k^3}{m}+\frac{k^3}{n}+\frac{k^3}{p}}\)

                 \(=\sqrt[3]{k^3\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)}\)

                \(=\sqrt[3]{k^3}=k\)

suy ra: đpcm

bài này ở trong Sách nâng cao và phát triển toán 9 tập 1 của ông Vũ Hữu Bình ý

Áp dụng BĐT Holder ta có:

\(\left(am^2+bn^2+cp^2\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\)

\(\ge\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)^3\)

Suy ra ĐPCM 

Dấu "=" xảy ra khi \(am^3=bn^3=cp^3\)

Đề: \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\sqrt{3}\) ???

*Ta chứng minh : \(x^4-x^3+2\ge x+1\forall x>0\)

\(\Leftrightarrow x^4-x^3-x+1\ge0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)\ge0\) ( đúng )

Do đó: \(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\) \(\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}=\sqrt{3}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)