\(1-\dfrac{1}{1+a}\ge\dfrac{2017}{b+2017}+\dfrac{2018}{c+2018}\ge2\sqrt{\dfrac{2017.2018}{\left(b+2017\right)\left(c+2018\right)}}\)

\(1-\dfrac{2017}{b+2017}\ge\dfrac{1}{1+a}+\dfrac{2018}{b+2018}\ge2\sqrt{\dfrac{2018}{\left(1+a\right)\left(b+2018\right)}}\)

\(1-\dfrac{2018}{c+2018}\ge\dfrac{1}{1+a}+\dfrac{2017}{b+2017}\ge2\sqrt{\dfrac{2017}{\left(1+a\right)\left(b+2017\right)}}\)

Nhân vế:

\(\dfrac{abc}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\ge\dfrac{8.2017.2018}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\)

\(\Rightarrow abc\ge8.2017.2018\)

pip install pygame

Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)

  \(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)

Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)

                         \(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

Cộng vế với vế ta có:

\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)

\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)

\(...\Leftrightarrow\dfrac{x+y+1}{6xy}=\dfrac{1}{6}\Leftrightarrow x+y+1=xy\Leftrightarrow\left(x-1\right)\left(y-1\right)=2\Leftrightarrow\left[{}\begin{matrix}x=3;y=2\\x=2;y=3\end{matrix}\right.\)

Mình biến đổi nhầm. Nhưng theo hướng đó bạn có thể làm cách khác.

Ta có: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\)

=>\(\frac{yz + zx + xy}{xyz}=0\)

=>xy+yz+xz=0

=>yz=-xy-xz; xy=-xz-yz; xz=-xy-yz

\(x^2 + 2yz = x^2 + yz + yz = x^2 - x(y + z) + yz\)

\(=x^2-xy-xz+yz\)

\(=x(x-y)-z(x-y)=(x-y)(x-z)\)

Chứng minh tương tự, ta sẽ có:

\(y^2 + 2zx = (y - z)(y - x)\)

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

Đặt \(A = \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy}\)

\(=\frac{1}{(x - y)(x - z)}+\frac{1}{(y - z)(y - x)}+\frac{1}{(z - x)(z - y)}\)

\(=\frac{-1}{(x - y)(z - x)}+\frac{-1}{(y - z)(x - y)}+\frac{-1}{(z - x)(y - z)}\)

\(=\frac{-(y - z) - (z - x) - (x - y)}{(x - y)(y - z)(z - x)}\)

=0

=>\(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}=0\)

\(\left(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)\)

\(=0\left(x^{2016}+y^{2017}+z^{2018}\right)\)

=0

=xy+yz+xz

Ta có:

\(P^2\)=\(\dfrac{x+y}{x+y-4034+2\sqrt{\left(x-2017\right)\left(y-2017\right)}}\)

\(P^2\)=\(\dfrac{x+y}{x+y-4034+2\sqrt{xy-2017\left(x+y\right)+2017^2}}\)

Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2017}\)

Suy ra xy=2017(x+y)

Suy ra \(P^2=\dfrac{x+y}{x+y-4034+2\sqrt{2017\left(x+y\right)-2017\left(x+y\right)+2017^2}}\)

\(P^2=\dfrac{x+y}{x+y-4034+2\sqrt{2017^2}}\)

\(P^2=\dfrac{x+y}{x+y-4034+4034}=\dfrac{x+y}{x+y}=1\)

Vậy P=1

Dark Bang SilentNam NguyễnNguyễn Huy Túlê thị hương giangMashiro ShiinaNgô Tấn ĐạtNguyễn Thanh HằngHà Nam Phan Đình

Ta có \(\sqrt{x}-\sqrt{x-1}< \dfrac{1}{100}\Leftrightarrow\dfrac{1}{\sqrt{x}+\sqrt{x-1}}< \dfrac{1}{100}\Leftrightarrow\sqrt{x}+\sqrt{x-1}>100\).

Đến đây dùng pp kẹp ta tìm được số nguyên dương x nhỏ nhất thỏa mãn là x = 2501.

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)

Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)