Question

1,Chứng minh rằng:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1990^2}< \dfrac{3}{4}\)

2,Chứng minh rằng:

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

Answer 1

Giúp mình với nhé

Answer 2

2) Mình nghĩ nên nhỏ hơn 3 thì dễ tính hơn... @@
Ta có :

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

\(\Rightarrow\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< \dfrac{x}{x}+\dfrac{y}{y}+\dfrac{z}{z}\\ \Rightarrow\dfrac{x+y+z}{x+y+z}< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< 1+1+1\\ \Rightarrow1< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< 3\)