Áp dụng cauchy-schwarz:

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+e}+\dfrac{d}{e+a}+\dfrac{e}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ce}+\dfrac{d^2}{ed+ad}+\dfrac{e^2}{ae+be}\ge\dfrac{\left(a+b+c+d\right)^2}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\)

Giờ chỉ cần chứng minh

\(ab+ac+ad+ae+bc+bd+be+cd+ce+de\le\dfrac{2}{5}\left(a+b+c+d+e\right)^2\)

\(\Leftrightarrow ab+ac+ad+ae+bc+bd+be+cd+ce+de\le2\left(a^2+b^2+c^2+d^2+e^2\right)\)

điều này hiển nhiên đúng theo AM-GM:

\(ab\le\dfrac{a^2+b^2}{2};ac\le\dfrac{a^2+c^2}{2};ad\le\dfrac{a^2+d^2}{2}...\)

Cứ vậy ta thu được đpcm .Dấu = xảy ra khi a=b=c=d=e

P/s: : ]

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

Ta có:

\(\dfrac{a}{f}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}.\dfrac{e}{f}=k^5=\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5\)

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=k\Rightarrow a=bk;b=ck;c=dk;d=ek\)

\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)

\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)

Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)

Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)

Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=t\) ta có:

\(\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{5d^4}{5e^4}=t^4\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(t^4=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\)

Mặt khác: \(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}=\dfrac{a}{e}=t.t.t.t=t^4\)

Ta có đpcm

a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)

\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)

\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)

BĐT mà ghi thiếu điều kiện thì chết rồi, vì số thực, số dương, số không âm nó hoạt động khác nhau lắm 

Bunhiacopxki: \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2+b^2\right)^4\)

\(\Rightarrow ac+bd\le\left(a^2+b^2\right)^2\)

Do đó:

\(\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^4}{ac}+\dfrac{b^4}{bd}\ge\dfrac{\left(a^2+b^2\right)^2}{ac+bd}\ge\dfrac{\left(a^2+b^2\right)^2}{\left(a^2+b^2\right)^2}=1\) (đpcm)

Đề bài sai: phản ví dụ:

\(a=b=-1\) ; \(c=d=2\)

Khi đó: \(c^2+d^2=\left(a^2+b^2\right)^3\) nhưng \(\dfrac{a^3}{c}+\dfrac{b^3}{d}=-1< 1\)

Bạn à tôi chịu

thì nó khó thiệt mà

Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)

Ta có: \(\dfrac{a^4}{b^4}=\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{a}{b}\)

\(=\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{d}\cdot\dfrac{e}{f}\)

\(=\dfrac{a}{f}\)