Dùng Shwarz là ra ngay nhé !

Ta có : \(1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\right)>1-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)=1-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1-\left(1-\dfrac{1}{100}\right)=1-1+\dfrac{1}{100}=\dfrac{1}{100}\)

Vậy \(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-.......-\dfrac{1}{100^2}>\dfrac{1}{100}\)

Xét \(\dfrac{x}{x+y+z+t}< \dfrac{x}{x+y+z}< \dfrac{x}{x+y}\)

\(\dfrac{y}{x+y+t+z}< \dfrac{y}{x+y+t}< \dfrac{y}{x+y}\)

\(\dfrac{z}{y+z+t+x}< \dfrac{z}{y+z+t}< \dfrac{z}{z+t}\)

\(\dfrac{t}{x+z+t+y}< \dfrac{t}{x+z+t}< \dfrac{t}{z+t}\)

Cộng cả ba vế , ta được :

\(\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}< \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}< \dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{z}{z+t}+\dfrac{t}{z+t}\)

\(\Rightarrow\dfrac{x+y+z+t}{x+y+z+t}< M< \dfrac{x+y}{x+y}+\dfrac{z+t}{z+t}\)

\(\Rightarrow1< M< 2\)

Vậy M không phải số tự nhiên

+ \(P=\frac{x}{y^2+1}+\frac{1}{y^2+1}+\frac{y}{z^2+1}+\frac{1}{z^2+1}+\frac{z}{x^2+1}+\frac{1}{x^2+1}\)

+ \(\frac{1}{x^2+1}=\frac{x^2+1-x^2}{x^2+1}=1-\frac{x^2}{x^2+1}\)

+ \(x^2+1\ge2x\forall x\)

\(\Rightarrow\frac{x^2}{x^2+1}\le\frac{x^2}{2x}=\frac{x}{2}\)

\(\Rightarrow-\frac{x^2}{x^2+1}\ge-\frac{x}{2}\)

\(\Rightarrow\frac{1}{x^2+1}\ge1-\frac{x}{2}\)

Dấu "=" xảy ra <=> x = 1

+ Tương tự ta cm đc :

\(\frac{1}{y^2+1}\ge1-\frac{y}{2}\). Dấu "=" xảy ra <=> y = 1

\(\frac{1}{z^2+1}\ge1-\frac{z}{2}\). Dấu "=" xảy ra <=> z = 1

Do đó : \(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\ge3-\left(\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\right)\)

\(\Rightarrow\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\ge3-\frac{3}{2}=\frac{3}{2}\) (1)

Dấu "=" xảy ra <=> x = y = z = 1.

+ \(\frac{x}{y^2+1}=\frac{x\left(y^2+1\right)-xy^2}{y^2+1}=x-\frac{xy^2}{y^2+1}\)

\(\Rightarrow\frac{x}{y^2+1}\ge x-\frac{xy^2}{2y}=x-\frac{xy}{2}\) ( do \(y^2+1\ge2y\forall y\) )

Dấu "=" xảy ra <=> y = 1.

Tương tự : \(\frac{y}{z^2+1}\ge y-\frac{yz}{2}\). Dấu "=" xảy ra <=> z = 1.

\(\frac{z}{x^2+1}\ge z-\frac{zx}{2}\). Dấu "=" xảy ra <=> x = 1.

Do đó : \(\frac{x}{y^2+1}+\frac{y}{z^2+1}+\frac{z}{x^2+1}\ge\left(x+y+z\right)-\frac{xy+yz+zx}{2}\)

\(\Rightarrow\frac{x}{y^2+1}+\frac{y}{z^2+1}+\frac{z}{x^2+1}\ge3-\frac{\frac{\left(x+y+z\right)^2}{3}}{2}\)

( do \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\) )

\(\Rightarrow\frac{x}{y^2+1}+\frac{y}{z^2+1}+\frac{z}{x^2+1}\ge3-\frac{3}{2}=\frac{3}{2}\) (2)

Dấu "=" xảy ra <=> x = y = z = 1.

Từ (1) và (2) suy ra

\(P\ge\frac{3}{2}+\frac{3}{2}=3\)

P = 3 \(\Leftrightarrow x=y=z=1\)

Vậy Min P = 3 \(\Leftrightarrow x=y=z=1\).

x,y,z>0 nha

\(M=\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+\dfrac{1}{16z^2}=\dfrac{1}{16}\left(\dfrac{1}{x^2}+\dfrac{2^2}{y^2}+\dfrac{4^2}{z^2}\right)\)

\(\Rightarrow M\ge\dfrac{1}{16}.\dfrac{\left(1+2+4\right)^2}{\left(x^2+y^2+z^2\right)}=\dfrac{49}{16}\)

\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\dfrac{1}{x^2}=\dfrac{2}{y^2}=\dfrac{4}{z^2}\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{7}\\y^2=\dfrac{2}{7}\\z^2=\dfrac{4}{7}\end{matrix}\right.\)

TH1: \(x+y+z+t=0\)

\(P=\left(1+\dfrac{x+y}{z+t}\right)^{2023}+\left(1+\dfrac{y+z}{x+t}\right)^{2023}+\left(1+\dfrac{z+t}{x+y}\right)^{2023}+\left(1+\dfrac{t+x}{y+z}\right)^{2023}\)

\(=\left(\dfrac{x+y+z+t}{z+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+y}\right)^{2023}+\left(\dfrac{x+y+z+t}{y+z}\right)^{2023}\)

\(=0+0+0+0=0\) là số nguyên (thỏa mãn)

TH2: \(x+y+z+t\ne0\), áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2023x+y+z+t}=\dfrac{y}{x+2023y+z+t}=\dfrac{z}{x+y+2023z+t}+\dfrac{t}{x+y+z+2023t}\)

\(=\dfrac{x+y+z+t}{\left(2023x+y+z+t\right)+\left(x+2023y+z+t\right)+\left(x+y+2023z+t\right)+\left(x+y+z+2023t\right)}\)

\(=\dfrac{x+y+z+t}{2026\left(x+y+z+t\right)}=\dfrac{1}{2026}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2023x+y+z+t}=\dfrac{1}{2026}\\\dfrac{y}{x+2023y+z+t}=\dfrac{1}{2026}\\\dfrac{z}{x+y+2023z+t}=\dfrac{1}{2026}\\\dfrac{t}{x+y+z+2023t}=\dfrac{1}{2026}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2026x=2023x+y+z+t\\2026y=x+2023y+z+t\\2026z=x+y+2023z+t\\2026t=x+y+z+2023t\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4x=x+y+z+t\\4y=x+y+z+t\\4z=x+y+z+t\\4t=x+y+z+t\end{matrix}\right.\)

\(\Rightarrow4x=4y=4z=4t\) (vì đều bằng \(x+y+z+t\))

\(\Rightarrow x=y=z=t\)

Do đó:

\(P=\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}\)

\(=2^{2023}+2^{2023}+2^{2023}+2^{2023}\)

\(=4.2^{2023}=2^{2025}\in Z\)

Em kiểm tra lại đề, 2 ngoặc cuối bị giống nhau, chắc em ghi nhầm

Chứng minh biểu thức thế nào em?