Question

[Dương Phan Khánh Vũ]

cho a,b,c là độ dài 3 cạnh tam giác. CMR a(b-c)²+b(c-a)²+c(a+b)²>a³+b³+c³

Answer 1

[Nguyễn Tấn Phát]

Ta thấy trong tam giác tổng độ dài hai cạnh luôn lớn hơn cạnh còn lại

Ta có: \(a+b>c\)

\(\Rightarrow\left(a+b\right)^2>c^2\)

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

Tương tự: 

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

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

do đó \(a\left(b+c\right)^2+b\left(a+c\right)^2+c\left(a+b\right)^2>a^3+b^3+c^3\left(ĐPCM\right)\)

Answer 2

[๖ۣۜҨž ♫ ℱ¡ɗℰ£¡ɑ๖²⁴ʱ]

Ta có:

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

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

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

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

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

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

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

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

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

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

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

vì a, b, c là cạnh của 1 tam giác

\(\Rightarrow\hept{\begin{cases}a+b-c>0\\c-a+b>0\\c+a-b>0\end{cases}}\)

\(\Rightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a+b\right)^2-a^3-b^3-c^3>0\)

\(\Rightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a+b\right)^2>a^3+b^3+c^3\)\(\left(đpcm\right)\)