\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}\)\(=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

\(\Rightarrow\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

=>Đpcm

đoàn cẩm lý sai rồi

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

\(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)

\(\frac{a+b^3}{c+d^3}=\frac{bk+b^3}{dk+d^3}\)

Đề bài sai nhé bạn

Xét: \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\)

\(\Leftrightarrow a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{cd^2}{c^2+d^2}+d-\frac{da^2}{d^2+a^2}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}a^2+b^2\ge2\sqrt{a^2b^2}=2ab\\b^2+c^2\ge2\sqrt{b^2c^2}=2bc\\c^2+d^2\ge2\sqrt{c^2d^2}=2cd\\d^2+a^2\ge2\sqrt{d^2a^2}=2da\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{ab^2}{a^2+b^2}\le\frac{ab^2}{2ab}=\frac{b}{2}\\\frac{bc^2}{b^2+c^2}\le\frac{bc^2}{2bc}=\frac{c}{2}\\\frac{cd^2}{c^2+d^2}\le\frac{cd^2}{2cd}=\frac{d}{2}\\\frac{da^2}{d^2+a^2}\le\frac{da^2}{2da}=\frac{a}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a-\frac{ab^2}{a^2+b^2}\ge a-\frac{b}{2}\\b-\frac{bc^2}{b^2+c^2}\ge b-\frac{c}{2}\\c-\frac{cd^2}{c^2+d^2}\ge c-\frac{d}{2}\\d-\frac{da^2}{d^2+a^2}\ge d-\frac{a}{2}\end{matrix}\right.\)

\(\Rightarrow a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{cd^2}{c^2+d^2}+d-\frac{da^2}{d^2+a^2}\ge a+b+c+d-\frac{a}{2}-\frac{b}{2}-\frac{c}{2}-\frac{d}{2}\)

\(\Rightarrow a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{cd^2}{c^2+d^2}+d-\frac{da^2}{d^2+a^2}\ge\frac{a+b+c+d}{2}\)

\(\Leftrightarrow\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\ge\frac{a+b+c+d}{2}\) ( đpcm )

Cách của bạn Minh dài quá mình xin làm cách ngắn hơn:

Đầu tiên ta chứng minh bổ đề:

\(\frac{x^3}{x^2+y^2}\ge\frac{2x-y}{2}\)

\(\Leftrightarrow2x^3-\left(x^2+y^2\right)\left(2x-y\right)\ge0\)

\(\Leftrightarrow y\left(y-x\right)^2\ge0\)(đúng)

Từ đó ta có: \(\left\{\begin{matrix}\frac{a^3}{a^2+b^2}\ge\frac{2a-b}{2}\\\frac{b^3}{b^2+c^2}\ge\frac{2b-c}{2}\\\frac{c^3}{c^2+d^2}\ge\frac{2c-d}{2}\\\frac{d^3}{d^2+a^2}\ge\frac{2d-a}{2}\end{matrix}\right.\)

Cộng 4 cái trên vế theo vế ta được

\(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\ge\frac{2a-b}{2}+\frac{2b-c}{2}+\frac{2c-d}{2}+\frac{2d-a}{2}=\frac{a+b+c+d}{2}\)

Áp dụng BĐT cauchy-schwarz :

\(VT=\frac{a^4}{ab+ac+ad}+\frac{b^4}{ab+bc+bd}+\frac{c^4}{cd+ac+bc}+\frac{d^4}{ad+bd+cd}\)

\(\ge\frac{\left(a^2+b^2+c^2+d^2\right)^2}{2\left(ab+ac+ad+bc+bd+cd\right)}\)

Mà \(3\left(a^2+b^2+c^2+d^2\right)\ge2\left(ab+ac+ad+bc+bd+cd\right)\)( dễ dàng chứng minh nó bằng AM-GM)

nên \(VT\ge\frac{a^2+b^2+c^2+d^2}{3}\)

Áp dụng BĐT AM-GM: \(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+d^2\ge2cd;d^2+a^2\ge2ad\)

\(\Rightarrow a^2+b^2+c^2+d^2\ge ab+bc+cd+da=1\)

do đó \(VT\ge\frac{1}{3}\)

Dấu''='' xảy ra khi \(a=b=c=d=\frac{1}{2}\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow a=b.k;b=c.k;c=d.k\)\(\Rightarrow a=k^3\)

\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{k^3.\left(b+c+d\right)^3}{b^3+c^3+d^3}=k^3\)

\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

=>đpcm

a/b=b/c=c/d

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vậy ..........