a/ Biến đổi tương đương:

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

\(\Leftrightarrow a^2c-a^2b+ab^2-ac^2+bc^2-b^2c\ge0\)

\(\Leftrightarrow a^2\left(c-b\right)-\left(ab+ac\right)\left(c-b\right)+bc\left(c-b\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a^2+bc-ab-ac\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a\left(a-b\right)-c\left(a-b\right)\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a-c\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(c-a\right)\left(b-a\right)\ge0\) luôn đúng do \(a\le b\le c\)

Vậy BĐT ban đầu đúng

Câu 2: Đề sai, cho \(a=b=c=1\Rightarrow3\ge6\) (sai)

Đề đúng phải là \(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

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

Câu 3: Không phải với mọi x; y với mọi \(x;y\) dương

Biến đổi tương đương do mẫu số vế phải dương nên ta được quyền nhân chéo:

\(\Leftrightarrow3x^3\ge\left(2x-y\right)\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow3x^3\ge2x^3+x^2y+xy^2-y^3\)

\(\Leftrightarrow x^3+y^3-x^2y-xy^2\ge0\)

\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (luôn đúng)

a: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3x^2y+y^3\)

\(=6x^2y+2y^3\)

\(=2y\left(3x^2+y^2\right)\)

a) https://hoc24.vn/hoi-dap/question/398481.html

b)

a² + b² + c² = ab + ac + bc

<=> 2a² + 2b² + 2c² = 2ac + 2ab + 2bc

<=> (a² - 2ac + c²) + (a² - 2ab + b²) + (b² - 2bc + c²) = 0

<=> (a - b)² + (a - c)² + (b - c)² = 0

<=> a = b = c

1. Ta có:

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

=> \(a^2y^2+b^2x^2=2axby\)

=> \(a^2y^2+b^2x^2-2axby=0\)

=> \(a^2y^2+b^2x^2-2aybx=0\)

=> \(\left(ay-bx\right)^2=0\)

Mà \(\left(ay-bx\right)^2\ge0\)

Dấu '' = '' xảy ra \(\Leftrightarrow\) \(ay-bx=0\)

\(\Leftrightarrow\) \(ay=bx\)

\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)

2. Ta có:

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

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

=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

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

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

Ta thấy:

\(\left(a-b\right)^2\ge0\); \(\left(a-c\right)^2\ge0\); \(\left(b-c\right)^2\ge0\)

=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

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

Dấu '' = '' xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\)

\(\Leftrightarrow\) a = b = c

\(\left(ab+bc+ca\right)^2=a^2b^2+2ab^2c+2a^2bc+b^2c^2+2c^2ba+c^2a^2=\left(a^2b^2+b^2c^2+c^2a^2\right)+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2\)

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