Bài làm :

Ta có :

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Leftrightarrow2ab+2bc+2ac=0\)

\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)

\(\Leftrightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)

\(\Leftrightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\left(1\right)\)

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

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

Thay (1) vào (2) ; ta được :

\(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

=> Điều phải chứng minh

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

\(\Leftrightarrow2ab+2ac+2bc=0\)

\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow ab+ac+bc=0\)

Ta lại có giả sử

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}=\frac{3}{abc}\)

\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=3\)

\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3=3.a^2b^2c^2\)

\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3-3.a^2b^2c^2=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)^3-3ca\left(ab+bc\right)\left(ab+bc+ac\right)-3ab^3c\left(-ac\right)-3a^2b^2c^2=0\)

\(\Leftrightarrow0+3a^2b^2c^2-3a^2b^2c^2+0=0\)

\(\Leftrightarrow0=0\left(lđ\right)\)

Vậy bất đẳng thức được chứng minh 

ta cm bài toán phụ nếu a+b+c=0 thì \(a^3+b^3+c^3=3abc\)

Thật vậy ta có \(a+b=-c\)=>(a+b)³=-c³<=> a³+b³+3ab(a+b)=-c³

=> a³+b³+c³=-3ab(-c)=3abc                (a+b=-c)

trở về bài toán ta có\(\left(a+b+c\right)^2=a^2+b^2+c^2\)=>\(ab+bc+ca=0\)

=>\(\frac{ab+bc+ca}{abc}=0\)=>\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

ÁP DỤNG BÀI TOÁN PHỤ ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) =>\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\frac{1}{a}\frac{1}{b}\frac{1}{c}=\frac{3}{abc}\)

Chứng minh bất đẳng thức \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)

Có: \(\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\left(\sqrt{x}^2+\sqrt{y}^2+\sqrt{z}^2\right)\ge\left(a+b+c\right)^2\) (Bunyakovsky)

\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)

abc = 1 => a^2.b^2.c^2 = 1

\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2c^2}{a^3\left(b+c\right)}+\frac{a^2b^2c^2}{b^3\left(c+a\right)}+\frac{a^2b^2c^2}{c^3\left(a+b\right)}\)
\(=\frac{\left(bc\right)^2}{ab+ac}+\frac{\left(ac\right)^2}{bc+ba}+\frac{\left(ab\right)^2}{ca+cb}\ge\frac{\left(ab+ac+bc\right)^2}{2\left(ab+ac+bc\right)}=\frac{\left(ab+ac+bc\right)}{2}\)
\(\ge\frac{3\sqrt[3]{ab.ac.bc}}{2}\)(Cauchy) \(=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}a=b=c\\\frac{bc}{ab+ac}=\frac{ac}{bc+ba}+\frac{ab}{ca+cb}\Leftrightarrow\end{cases}a=b=c}\)

Mà abc=1 <=> a^3 = 1 <=> a=1 => b=c=a=1

https://diendantoanhoc.net/topic/80159-ch%E1%BB%A9ng-minh-frac1a2b3cfrac12a3bcfrac13bb2c-leqslant-frac316/

bạn tham khảo ở đây nhé

cách này ổn nè :

với a,b,c \(\in\)R và x,y,z > 0 . Áp dụng BĐT : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)

Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

Ta có : \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{\frac{1}{a^2}}{ab+bc}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\)

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)( vì abc = 1 )

Mà \(\left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\right)^3\ge\frac{1}{abc}\)hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)

\(\Rightarrow\)\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1

Câu 1:

\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)

\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow0=0\) Đúng (Đpcm)

​Áp dụng Bđt Cô si 3 số ta có:

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)

Dấu = khi a=b=c (Đpcm)

Câu 2

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)

Ta có:

\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)

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

\(=abc\cdot3\cdot\frac{1}{abc}=3\)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{array}\right.\)

\(\Rightarrow A=\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)

\(=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)

\(=\left(-\frac{c}{b}\right)\left(-\frac{a}{c}\right)\left(-\frac{b}{a}\right)=-1\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a-b=b-c=c-a=0\Leftrightarrow a=b=c\)

\(\Leftrightarrow A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Em lớp 8, mạn phép làm bài này ạ , có gì sai mong mn chỉ bảo :33

BĐT cần chứng minh 

\(\Leftrightarrow3+\frac{b^2+c^2}{a^2}+\frac{a^2+c^2}{b^2}+\frac{a^2+b^2}{c^2}\ge3+\frac{2\left(a^3+b^3+c^3\right)}{abc}\)

\(\Leftrightarrow\frac{b^2+c^2}{a^2}+\frac{a^2+c^2}{b^2}+\frac{a^2+b^2}{c^2}\ge\frac{2\left(a^3+b^3+c^3\right)}{abc}\) (1)

Ta có : \(VT\left(1\right)\ge\frac{2bc}{a^2}+\frac{2ac}{b^2}+\frac{2ab}{c^2}\ge3\sqrt[3]{\frac{8\left(abc\right)^2}{\left(abc\right)^2}=6}\)

\(VT\left(1\right)\ge\frac{2.3\sqrt[3]{\left(abc\right)^3}}{abc}=6\)

Do đó (1) đúng. (đpcm)