Question

Cho các số x, y cùng dấu. CM rằng:

a) \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

b) \(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge0\)

c)\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{x^2}{y^2}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)

Answer 1

a)\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2+y^2-2xy}{xy}=\dfrac{\left(x-y\right)^2}{xy}\)\(\ge0\)

Vậy \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

b) ta có: A=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

A\(\ge\)\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\)

=\(\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)

Answer 2

Answer 3

c) Từ câu b suy ra:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)\ge0\)

Vì \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)(câu a)

Nên:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)

Answer 4

Vì \(x;y\) cùng dấu \(\Rightarrow\dfrac{x}{y};\dfrac{y}{x}>0\)

Áp dụng Bất đẳng thức Cauchy cho các cặp số dương \(\left(\dfrac{x}{y};\dfrac{y}{x}\right);\left(\dfrac{x^2}{y^2};\dfrac{y^2}{x^2}\right);\left(\dfrac{x^4}{y^4};\dfrac{y^4}{x^4}\right)\)

\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt[]{\dfrac{x}{y}.\dfrac{y}{x}}=2\left(1\right)\) \(\rightarrow câu.a\)

\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\ge2\sqrt[]{\dfrac{x^2}{y^2}.\dfrac{y^2}{x^2}}=2\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge0\rightarrow câu.b\)

\(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\ge2\sqrt[]{\dfrac{x^4}{y^4}.\dfrac{y^4}{x^4}}=2\left(3\right)\)

\(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\rightarrow câu.c\)