\(\left(1-\frac14\right)\left(1-\frac19\right)\left(1-\frac{1}{16}\right)\cdot\ldots\cdot\left(1-\frac{1}{\left(n+1\right)^2}\right)\)

\(=\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\cdot\ldots\cdot\left(1-\frac{1}{n+1}\right)\left(1+\frac12\right)\left(1+\frac13\right)\left(1+\frac14\right)\cdot\ldots\cdot\left(1+\frac{1}{n+1}\right)\)

\(=\frac12\cdot\frac23\cdot\frac34\cdot\ldots\cdot\frac{n}{n+1}\cdot\frac32\cdot\frac43\cdot\frac54\cdot\ldots\cdot\frac{n+2}{n+1}\)

\(=\frac{1}{n+1}\cdot\frac{n+2}{2}=\frac{n+2}{2\left(n+1\right)}\)

1-1=

1+1=?

Nếu không phải là đố mẹo thì = 10.

Nếu là đố mẹo thì...mình không biết!

Ta có: 

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{1}{a}\ge\dfrac{2b-1}{2b+1}+\dfrac{3c-1}{3c+2}\ge2\sqrt{\dfrac{\left(2b-1\right)\left(3c-1\right)}{\left(2b+1\right)\left(3c+2\right)}}\left(1\right)\)

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{2}{2b+1}\ge\dfrac{a-1}{a}+\dfrac{3c-1}{3c+2}\ge2\sqrt{\dfrac{\left(a-1\right)\left(3c-1\right)}{a\left(3c+2\right)}}\left(2\right)\)

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{3}{3c+2}\ge\dfrac{a-1}{a}+\dfrac{2b-1}{2b+1}\ge2\sqrt{\dfrac{\left(a-1\right)\left(2b-1\right)}{a\left(2b+1\right)}}\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow6\ge8\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)

\(\Rightarrow P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\le\dfrac{3}{4}\)

\(\Rightarrow P_{max}=\dfrac{3}{4}\) đạt tại \(a=\dfrac{3}{2};b=1;c=\dfrac{5}{6}\)

BĐT sai khi \(a;b;c\) thuộc \(\left(0;1\right)\) và \(a;b;c\) không bằng nhau

Đặt \(\left(a;b;c\right)=\left(\dfrac{yz}{x^2};\dfrac{zx}{y^2};\dfrac{xy}{z^2}\right)\)

BĐT cần chứng minh tương đương:

\(\dfrac{x^4}{x^4+x^2yz+y^2z^2}+\dfrac{y^4}{y^4+xy^2z+z^2x^2}+\dfrac{z^4}{z^4+xyz^2+x^2y^2}\ge1\)

Ta có:

\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+xyz\left(x+y+z\right)+x^2y^2+y^2z^2+z^2x^2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+xyz\left(x+y+z\right)+x^2y^2+y^2z^2+z^2x^2}\ge1\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)\) (hiển nhiên đúng)

Dấu "=" xảy ra khi \(a=b=c=1\)