cho M=x/(x+y+z)+y/(x+y+t)+z/(y+z+t)+t/(x+z+t) voi x, y, z, t thuoc N*
CM M^10<1025
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\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
\(=\frac{x+y+z+t}{y+z+t+z+t+x+t+x+y+x+y+z}=\frac{x+y+z+t}{3x+3y+3z+3t}\)
\(=\frac{x+y+z+t}{3\left(x+y+z+t\right)}=\frac{1}{3}\)
\(\Rightarrow x=y=z=t\)
\(=\frac{x+y}{z+t}+\frac{y+z}{x+t}+\frac{z+t}{x+y}+\frac{x+t}{x+z}=\frac{x+x}{x+x}+\frac{y+y}{y+y}+\frac{z+z}{z+z}+\frac{t+t}{t+t}=4\)
Ta có :
\(\frac{x}{x+y+z}< 1\)\(\Rightarrow\frac{x}{x+y+z+t}< \frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}\)( 1 )
\(\frac{y}{x+z+t}< 1\)\(\Rightarrow\frac{y}{x+y+z+t}< \frac{y}{x+z+t}< \frac{x+y}{x+y+z+t}\)( 2 )
\(\frac{z}{y+z+t}< 1\)\(\Rightarrow\frac{z}{x+y+z+t}< \frac{z}{y+z+t}< \frac{y+z}{x+y+z+t}\)( 3 )
\(\frac{t}{x+z+t}< 1\)\(\Rightarrow\frac{t}{x+y+z+t}< \frac{t}{x+z+t}< \frac{z+t}{x+y+z+t}\)( 4 )
cộng ( 1 ) , ( 2 ) , ( 3 ) và ( 4 ) ta được :
\(\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}+\frac{t}{x+y+z+t}\)
\(< \frac{x}{x+y+z}+\frac{y}{x+z+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}\)
\(< \frac{x+t}{x+y+z+t}+\frac{x+y}{x+y+z+t}+\frac{y+z}{x+y+z+t}+\frac{z+t}{x+y+z+t}\)
\(\Leftrightarrow1< \frac{x}{x+y+z}+\frac{y}{x+z+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}< 2\)
Vậy M không là số tự nhiên
Ta chứng minh tính chất \(\frac{a}{b}< 1\) suy ra \(\frac{a+m}{b+m}>\frac{a}{b}\)
Ta có \(1-\frac{a}{b}=\frac{b-a}{b}\)
\(1-\frac{a+m}{b+m}=\frac{b-a}{b+m}\)
Vì \(\frac{b-a}{b}>\frac{b-a}{b+m}=>\frac{a}{b}< \frac{a+m}{b+m}\)
Áp dụng thính chất trên ta có
\(M< \frac{x+t}{x+y+z+t}+\frac{y+z}{x+y+t+z}+\frac{z+x}{y+z+t+x}+\frac{t+y}{x+z+t+y}\)
=> M < 2 => M10 <210=1024 <1025
Vậy M10 <1025
\(M< \frac{x+t}{x+y+z+t}+\frac{y+z}{x+y+z+t}+\frac{x+z}{x+y+z+t}+\frac{y+t}{x+y+z+t}\)
\(\Rightarrow M< \frac{\left(x+t\right)+\left(y+z\right)+\left(x+z\right)+\left(y+t\right)}{x+y+z+t}\)
\(\Rightarrow M< \frac{2\left(x+y+z+t\right)}{x+y+z+t}\Rightarrow M< 2\)
\(\Rightarrow M^{10}< 2^{10}\Rightarrow M^{10}< 1024\Rightarrow M^{10}< 1025\)
thanks Pham Van Hung