Ta có \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Để tổng trên chia hết cho 81 thì \(\left(a-b\right)\left(b-c\right)\left(c-a\right)⋮27\)

Mà \(a+b+c=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Bài toán trở thành: Cho \(x+y+z=\left(x-y\right)\left(y-z\right)\left(z-x\right)\). CMR: \(x+y+z⋮27\) - Hoc24

wow

wow, chắc xu học lớp 9

\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(b+c\right)}{64}}=\dfrac{3a}{4}\)

Tương tự:

\(\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}+\dfrac{c+a}{8}+\dfrac{a+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(VT+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)

Đặt BĐT cần c/m là A

Dự đoán đẳng thức xảy ra khi a = b = c

Áp dụng BĐT Cauchy cho 3 số không âm:

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\)

\(\ge3\sqrt[3]{\frac{a^3}{\left(a+b\right)\left(a+c\right)}.\frac{a+b}{8}.\frac{a+c}{8}}=\frac{3a}{4}\)

\(\frac{b^3}{\left(b+c\right)\left(b+a\right)}+\frac{b+c}{8}+\frac{b+a}{8}\)

\(\ge3\sqrt[3]{\frac{b^3}{\left(b+c\right)\left(b+a\right)}.\frac{b+c}{8}.\frac{b+a}{8}}=\frac{3b}{4}\)

\(\frac{c^3}{\left(c+a\right)\left(c+b\right)}+\frac{c+a}{8}+\frac{c+b}{8}\)

\(\ge3\sqrt[3]{\frac{c^3}{\left(c+a\right)\left(c+b\right)}.\frac{c+a}{8}.\frac{c+b}{8}}=\frac{3c}{4}\)

Cộng từng vế của các BĐT trên, ta được:

\(A+\frac{2\left(a+b+c\right)}{4}\ge\frac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow A\ge\frac{3}{4}\)

(Dấu "="\(\Leftrightarrow a=b=c\))

Dấu >= hay <= vậy bạn? Bạn xem lại đề.

\(\left(a+b+c\right)^3\)

\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

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

\(=a^3+b^3+c^3+3a^2b+3b^2a+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)

\(=a^3+b^3+c^3+3\left(a^2b+b^2a+a^2c+2abc+b^2c+ac^2+bc^2\right)\)

\(=a^3+b^3+c^3+3\left[ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)\right]\)

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

\(\left(đpcm\right)\)

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

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

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

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

...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

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

#)Giải :

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

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

\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b-c\right)\left(b^2-bc+c^2\right)\)

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

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

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

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

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

\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)

\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\Rightarrowđpcm\)

\(VT\ge\sum\left(\dfrac{a^3}{2a+b+c}\right)=\sum\left(\dfrac{a^3}{\sum a+a}\right)=\sum\dfrac{a^3}{3+a}\)

Ta có BĐT phụ :

\(\dfrac{a^3}{a+3}\ge\dfrac{11a-7}{16}\)(*)

\(\Leftrightarrow\left(16a+21\right)\left(a-1\right)^2\ge0\) (luôn đúng với mọi a>0)

Áp dụng BĐT (*) ta có :

\(\sum\dfrac{a^3}{3+a}\ge\dfrac{11\sum a-21}{16}=\dfrac{33-21}{16}=\dfrac{12}{16}=\dfrac{3}{4}\)

Bạn xem lời giải ở đây nhé https://olm.vn/hoi-dap/question/960694.html

Another way CLICK HERE

Ta biến đổi: \(4\left(a^3+b^3\right)-\left(a+b\right)^3+4\left(b^3+c^3\right)-\left(b+c\right)^3+4\left(c^3+a^3\right)-\left(c+a\right)^3\ge0\)

Xét: \(4\left(a^3+b^3\right)-\left(a+b\right)^3=\left(a+b\right)\left[4\left(a^2-ab+b^2\right)-\left(a+b\right)^2\right]\)

\(=3\left(x+b\right)\left(a-b\right)^2\ge0\)

Tương tự với: \(4\left(b^3+c^3\right)-\left(b+c\right)^3\) và \(4\left(c^3+a^3\right)-\left(c+a\right)^3\)

Ta suy ra đpcm.

Đẳng thức xảy ra \(\Leftrightarrow a=b=c\)