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 

Bài 1:

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

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

Theo tính chất dãy tỉ số bằng nhau, ta có:

a/b = b/c = c/d = (a + b + c)/(b + c + d)

--> ((a + b + c)/(b + c + d))^3 = a^3/b^3

Cần chứng minh:

a^3/b^3 = a/d

<=> a^3/b^3 = a^3/(a^2.d)

--> b^3 = a^2.d

Mà ad = bc (do a/b = c/d)

--> b^3 = abc

<=> b^2 = ac (luôn đúng do a/b = b/c)

--> đpcm

Bài 1:

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

Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)

\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

Ta có đpcm.

Bài 2:

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

Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)

Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.

Bài 3:

a) Sửa điều kiện: \(\frac{a}{b}=\frac{c}{d}\neq -1\)

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

Theo đkđb thì \(k\neq -1\) nên \(k^3+1\neq 0\); \(k+1\neq 0\)

Ta có: \(\frac{a^3+b^3}{c^3+d^3}=\frac{(bk)^3+b^3}{(dk)^3+d^3}=\frac{b^3(k^3+1)}{d^3(k^3+1)}=\frac{b^3}{d^3}\)

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

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

b)

Đặt \(\frac{a}{b}=k; \frac{c}{d}=t\Rightarrow a=bk; c=dt\)

Ta cần cm \(k=t\)

Khi đó:

\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b(2k+13)}{b(3k-7)}=\frac{2k+13}{3k-7}\)

\(\frac{2c+13d}{3c-7d}=\frac{2dt+13d}{3dt-7d}=\frac{d(2t+13)}{d(3t-7)}=\frac{2t+13}{3t-7}\)

Vì \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\Rightarrow \frac{2k+13}{3k-7}=\frac{2t+13}{3t-7}\)

\(\Rightarrow (2k+13)(3t-7)=(2t+13)(3k-7)\)

\(-14k+39t=-14t+39k\Rightarrow k=t\)

Ta có đpcm.

Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)

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

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

\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)

\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).

ĐTXR \(\Leftrightarrow a=b=c=1\)

Theo đề bài ta có :

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

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

Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) ( 1 )

Theo tính chất dãy tỉ số bằng nhau ta có :

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

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

Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )

Từ ( 2 ) , ( 3 ) 

 = > \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( đpcm )

Đặt a/2019=b/2021=c/2023=k

=>a=2019k; b=2021k; c=2023k

(a-c)^2/4=(2023k-2019k)^2/4=(4k)^2/4=4k^2

(a-b)(b-c)=(2019k-2021k)(2021k-2023k)=4k^2

=>(a-c)^2/4=(a-b)(b-c)