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 \(a=b=c\)

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

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 "="\(\Leftrightarrow a=b=c\))

\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)

Tương tự:

\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)

\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)

Cộng vế:

\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

chứng minh điều j bn

\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)

Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)

\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)

Cộng vế:

\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)

\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

chết đăng nhầm sogy nha