Chứng minh : a³ + b³ + c³ = 3abc \(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(tm\right)\\a=b=c\left(loai\right)\end{cases}}\)

Rút gọn P

\(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)}{abc}\)

Xét : ab(a-b) + bc(b-c) + ac(c-a) = ab[-(b-c)-(c-a)] + bc(b-c) + ac(c-a)

= (b-c)(bc-ab) + (c-a)(ac-ab) = b(b-c)(c-a) + a(c-a)(c-b) = (c-a)(c-b)(a-b)

\(\Rightarrow P=\frac{\left(c-a\right)\left(c-b\right)\left(a-b\right)}{abc}\)

Rút gọn Q

Đặt a - b = z ; b-c = x ; c - a = y

\(\Rightarrow\)x- y = a + b - 2c = -c - 2c = -3c              ( do a + b + c = 0 )

y - z = -3a ; z - x = -3b

\(\Rightarrow\)\(-3Q=\frac{\left(y-z\right)}{x}+\frac{\left(z-x\right)}{y}+\frac{\left(x-y\right)}{z}\)

Làm tương tự như rút gọn P, ta có : 

\(-3Q=\frac{\left(x-y\right)\left(z-y\right)\left(z-x\right)}{xyz}=\frac{-\left(-3a\right)\left(-3b\right)\left(-3c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{27abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-27abc}{\left(a-b\right)\left(c-b\right)\left(c-a\right)}\)

\(\Rightarrow Q=\frac{9abc}{\left(a-b\right)\left(c-b\right)\left(c-a\right)}\)

\(\Rightarrow PQ=9\)

Bài 1: \(a+\frac{1}{b\left(a-b\right)}=\left(a-b\right)+b+\frac{1}{b\left(a-b\right)}\)

Áp dụng BĐT Cauchy cho 3 số dương ta thu được đpcm (mình làm ở đâu đó rồi mà:)

Dấu "=" xảy ra khi a =2; b =1 (tự giải ra)

Bài 2: Thêm đk a,b,c >0.

Theo BĐT Cauchy \(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{c^2}}=\frac{2a}{c}\). Tương tự với hai cặp còn lại và cộng theo vế ròi 6chia cho 2 hai có đpcm.

Bài 3: Nó sao sao ấy ta?

Ta có:

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)

\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)

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

Ta có : \(b^2=ac\) 

\(\Rightarrow\frac{a}{b}=\frac{b}{c}\) (1) 

\(c^2=bd\) 

\(\Rightarrow\frac{b}{c}=\frac{c}{d}\) (2)

Từ (1) và (2) suy ra : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) 

\(\Rightarrow\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\) , \(\frac{b}{c}.\frac{b}{c}.\frac{b}{c}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\) và \(\frac{c}{d}.\frac{c}{d}.\frac{c}{d}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\)

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

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

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

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

 Ta có:

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)

\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)

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

\(ADTCDTSBN,\)ta có:

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

Lại có:\(\frac{a^3}{b^3}=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)

25361

Ta có : \(\frac{\frac{\left(a-b\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}\)

\(=\frac{\frac{\left(\sqrt{a}-\sqrt{b}\right)^3\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}+2a\sqrt{a}-b\sqrt{b}}{\sqrt{a}^3-\sqrt{b}^3}+\frac{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{-\left(a-b\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^3+2a\sqrt{a}-b\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{-\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}+2a\sqrt{a}-b\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{3a\sqrt{b}+3\sqrt{a}b+3a\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{3\sqrt{a}\left(\sqrt{ab}+b+a\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=-\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}=0\)

Vậy ...

\(\left(a+b+c\right)^2=a^2+b^2+c^2 \Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)

<=> \(ab+bc+ac=0\Leftrightarrow\frac{ab+ac+bc}{abc}=0\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\frac{1}{c^3}\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+3.\frac{1}{a^2}.\frac{1}{b}+3.\frac{1}{a}.\frac{1}{b^2}=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{-1}{c}\right)=0\Leftrightarrow\)dpcm