soit `a , b ` deux réels strictement positifs tels que ` a <= b `

Montrer que ` a <= 2/{1/a+1/b} <= sqrt(ab) <= {a+b}/2<={sqrt(a^2+b^2)}/2 <= b `

soit `x, y ` deux réels strictement positifs tels que `x <= y `

1 Montrons que : ` x <= 2/(1/x+1/y) `

on a ` 2/(1/x+1/y) - x = 2/((x+y)/(xy)) -x `

` = (2xy)/(x+y) -x `

` = (2xy -x^2-y^2)/(x+y) `

` = (xy -x^2)/(x+y)`

`= (x(y-x))/(x+y)`

Puisque ` x<= y ` et ` x > 0 , y > 0 `

`=> (x(y-x))/(x+y) >= 0 `

`=> 2/(1/x+1/y) - x >= 0 `

`=> x <= 2/(1/x+1/y) `

2 Montrons que : ` 2/(1/x+1/y) <= sqrt(xy) `

on a ` 2sqrt(xy) <= (x+y ) ` car ` (sqrt(x) -sqrt(y))^2 >= 0 `

`=> 2/(x+y) <= 1/(sqrt(xy))`

`=> 2/(x+y) <= (sqrt(xy))/(xy)`

`=> (2xy)/(x+y) <= sqrt(xy) ` car ` xy > 0 `

`=> 2/(x/(xy) + y/(xy)) <= sqrt(xy) `

`=> 2/(1/x+1/y) <= sqrt(xy) `

`=> 2/(1/x+1/y) <= sqrt(xy) `

3 Montrons que : ` sqrt(xy) <= (x+y)/2 `

on a `(sqrt(x) -sqrt(y))^2 >= 0 `

`=> x+ y -2sqrt(xy) >= 0 `

`=> x+ y >= 2sqrt(xy) `

`=> (x+y)/2 >= sqrt(xy) `

`=> sqrt(xy) <= (x+y)/2 `

4 Montrons que : ` (x+y)/2 <= sqrt((x^2+y^2) /2) `

on a `(x+y)/2 <= sqrt((x^2+y^2) /2) `

`<=> ((x+y)/2)^2 <=( sqrt((x^2+y^2) /2))^2 ` car `(x+y)/2 > 0 ` et `sqrt((x^2+y^2) /2) > 0 `

`<=> (x+y)^2/4 <= (x^2+y^2)/2 `

`<=> (2x^2+2y^2 -x^2-y^2-2xy) /4 >= 0 `

`<=> (x^2+y^2-2xy)/4 >= 0 `

`<=> (x-y)^2 >= 0 `

or cette pour tout `x, y ` dans `R` on a `(x-y)^2 >= 0 `

alors ` (x+y)/2 <= sqrt((x^2+y^2) /2) `

5 Montrons que `sqrt((x^2+y^2)/2) <= y `

on a ` sqrt((x^2+y^2)/2) <= y `

`<=> (x^2+y^2)/2 <= y^2 ` car ` 0 < y ` et ` 0 < sqrt((x^2+y^2)/2) `

`<=> x^2 +y^2 <= 2y^2 `

`<=> x^2 <= y^2 `

`<=> x <= y ` car ` 0 < x ` et ` 0 < y `

or cette dernière inégalité est vraie

Alors `sqrt((x^2+y^2)/2) <= y `

Conclusion

Pour tout ` x > 0 ` , ` y > 0 ` tels que ` x <= y `

`x <= 2/(1/x+1/y) <= sqrt(xy) <= (x+y)/2 <= sqrt((x^2+y^2)/2) <= y `