Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)

\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)

=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)

b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)

=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

Lời giải:

Ta có: \(a^2b+b^2c+c^2a\geq \frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)

\(\Leftrightarrow (a^2b+b^2c+c^2a)(1+2a^2b^2c^2)\geq 9a^2b^2c^2\)

\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)(*)\)

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Áp dụng BĐT AM-GM ta có:

\(a^2b+a^4b^3c^2+a^3b^2c^4\geq 3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)

\(b^2c+a^2b^4c^3+a^4b^3c^2\geq 3a^2b^3c^2\)

\(c^2a+a^3b^2c^4+a^2b^4c^3\geq 3a^2b^2c^3\)

Cộng theo vế:

\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)\)

Vậy $(*)$ đúng

Do đó ta có đpcm

Dấu bằng xảy ra khi $a=b=c=1$

Theo bài ra ta có : 

\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

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

Nếu a + b + c + d = 0

\(\Rightarrow\frac{0}{a}=\frac{0}{b}=\frac{0}{c}=\frac{0}{d}\)

\(\Rightarrow\orbr{\begin{cases}a=b=c=d\\a\ne b\ne c\ne d\end{cases}}\)(loại) 

Nếu a + b + c + d \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\)

=> a = b = c = d (đpcm)

Do a,b<1 => a^3<a^2<a<1 ; b^3<b^2<b<1 ; ta có :

(1-a^2)(1-b) => 1+a^2b>a^2+b

=> 1+a^2b>a^3+b^3 hay a^3+b^3 <1+a^2b

Tương tự : b^3+c^3 < 1+b^2;c^3+a^3<1+c^2a

=> 2a^3+2b^3+2c^3<3+a^2b+b^2c+c^2a