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Đặt a -b= x, b - c =y => c - a = -(x+y)
\(a^2b^2x+b^2c^2y-\left(x+y\right)c^2a^2\)
\(=x\left(a^2b^2-c^2a^2\right)+y\left(b^2c^2-c^2a^2\right)\)
\(=a^2\left(b+c\right)\left(a-b\right)\left(b-c\right)-c^2\left(a+b\right)\left(a-b\right)\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[a^2b+a^2c-c^2a-c^2b\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(a-c\right)\left(a+c\right)+ca\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)
P/s: có tính nhầm chỗ nào ko nhỉ?
\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=a^3-ab^2+a^2b-b^3+b^3-bc^2+b^2c-c^3+c^3-a^2c+ac^2-a^3\)
\(=-ab^2+a^2b-bc^2+b^2c-a^2c+ac^2\)
\(=\left(a^2b-ab^2\right)+\left(ac^2-bc^2\right)-\left(a^2c-b^2c\right)\)
\(=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)
\(=\left(a-b\right)\left[\left(ab-ac\right)+\left(c^2-bc\right)\right]\)
\(=\left(a-b\right)\left[a\left(b-c\right)+c\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)=\left(c-a\right)\left(c-b\right)\left(b-a\right)\)
a: \(=a^2b-ab^2+b^2c-bc^2+c^2a-ca^2\)
\(=a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac+b^2-bc+c^2\right)\)
b: \(=b^2c+bc^2+ac^2-a^2c-a^2b-ab^2\)
\(=b^2\left(c-a\right)+b\left(c^2-a^2\right)+ac\left(c-a\right)\)
\(=\left(c-a\right)\left(b^2+ac+b\left(c+a\right)\right)\)
\(=\left(c-a\right)\left(b^2+ac+bc+ba\right)\)
\(=\left(c-a\right)\left(b+c\right)\left(b+a\right)\)
\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)
\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)
Áp dụng BĐT Cosi ta có:
\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)
Từ (1)(2)(3) ta có:
\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)
Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)
Dấu "=" xảy ra <=> a=b=c=1
có cả mấy bất đẳng thức đó hả
bn viết công thức tổng quát ra cho mk vs
mk thanks
\(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left[\left(a-b\right)+\left(c-a\right)\right]+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)+c^2a^2\left(c-a\right)-b^2c^2\left(c-a\right)\)
\(=\left(a-b\right)b^2\left(a-c\right)\left(a+c\right)+\left(c-a\right)c^2\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(ab^2+cb^2-c^2a-c^2b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+ac+bc\right)\)
t làm bên h rồi mà? Làm quá lâu rồi luôn ấy! Đáp án y chang bạn Kid:v
Câu hỏi của Trần Minh Hiển - Toán lớp 9 (không biết AD đã fix lỗi ko dán link h vào olm chưa, nếu chưa ib t gửi full link, nhớ kèm theo link câu hỏi này là ok.)