Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\b=ck\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{b^3k^3+c^3k^3+d^3k^3}{b^3+c^3+d^3}=k^3\)

\(\dfrac{a}{d}=\dfrac{bk}{d}=\dfrac{ck^2}{d}=\dfrac{dk^3}{d}=k^3\)

Do đó: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)

Từ bài toán này (mà bạn đã hỏi cách đây vài bữa):

cho a,b,c>0. Chứng minh rằng: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\) - Hoc24

Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)

Do đó: \(VT\ge\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)

Lại có: \(\dfrac{a+b+c}{\sqrt[3]{abc}}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}=3\)

Đặt \(\dfrac{a+b+c}{\sqrt[3]{abc}}=x\ge3\Rightarrow VT\ge x+\dfrac{1}{x}=\dfrac{x}{9}+\dfrac{1}{x}+\dfrac{8x}{9}\ge2\sqrt{\dfrac{x}{9x}}+\dfrac{8}{9}.3=\dfrac{10}{3}\) (đpcm)

Dương hay không âm bạn?

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D 

\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)

Ta chứng minh bđt phụ \(x^2+y^2+z^2\ge xy+yz+zx\forall x,y,z>0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\Leftrightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\ge0\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)\(\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\left(1\right)\)

Áp dụng bđt Cô-si vào các số a,b,c dương :

\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}\cdot ab}=2\sqrt{a^4}=2a^2\)

Chứng minh tương tự ta được:

\(\dfrac{b^3}{c}+bc\ge2b^2;\dfrac{c^3}{a}+ca\ge2c^2\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\ge2ab+2bc+2ca\) (do áp dụng (1)) \(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(ab+bc+ca\right)-\left(ab+bc+ca\right)=ab+bc+ca\)

Dấu = xảy ra \(\Leftrightarrow a=b=c\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)

Vậy \(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\left(dpcm\right)\)

\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)

\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)

\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)

\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)

giả sử \(a>b>c>0\) thì ta có :

\(\dfrac{a^2}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^2}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\ge2\dfrac{a}{b}+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\)

\(=\dfrac{2a}{b}+\dfrac{c^3}{a^3}-\dfrac{c^2}{a^2}\ge0\)

làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)

\(\Rightarrow\left(đpcm\right)\)