\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=6\)

\(P=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)

\(P\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{xy+yz+zx}{3}=2\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt{2}\)

Bạn quy đồng làm từ từ là đc

Đặt \(M=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

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

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

\(=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) là số chính phương.

Vậy ta có đpcm.

Đặt \(THANG=ab\left(a+1\right)+bc\left(b+1\right)+ca\left(c+1\right)\) :v

Vì \(0\le a;b;c\le1\)\(\Rightarrow\left\{{}\begin{matrix}a^2\left(1-b\right)\le a\left(1-b\right)\\b^2\left(1-c\right)\le b\left(1-c\right)\\c^2\left(1-a\right)\le c\left(1-a\right)\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2-\left(a^2b+b^2c+c^2a\right)\le a+b+c-\left(ab+bc+ca\right)\)

\(\Rightarrow\left(a^2b+b^2c+c^2a\right)+\left(a+b+c\right)\ge a^2+b^2+c^2+ab+bc+ca\)

\(\Rightarrow\left(a^2b+b^2c+c^2a\right)+\left(ab+bc+ca\right)+\left(a+b+c\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Rightarrow THANG\ge\left(a+b+c\right)^2-\left(a+b+c\right)=\left(a+b+c\right)\left(a+b+c-1\right)\)

Vì \(a+b+c\ge2\) nên \(a+b+c-1\ge1\). Vậy \(THANG\ge2\cdot1=2\)

Đẳng thức xảy ra khi trong 3 số \(a;b;c\) có 2 số bằng 1 và một số bằng 0