| a | a | b | b | |||||
Wykaż, że nierówność | (1+ | ) + | (1+ | ) ≥ 4 jest spełniona dla wszystkich | ||||
| b | b | a | a |
| a | ||
c = | > 0 | |
| b |
| 1 | 1 | |||
c(1+c) + | (1 + | ) ≥ 4 | ||
| c | c |
| 1 | 1 | |||
c2 −2 + | + c − 2 + | ≥ 0 | ||
| c2 | c |
| 1 | √c | |||
(c − | )2 + (√c − | )2 ≥ 0 | ||
| c | c |
| a | a | b | b | |||||
R= | *(1+ | )+ | *(1+ | )−4= | ||||
| b | b | a | a |
| a | a2 | b | b2 | |||||
= | + | + | + | −4= grupujemy | ||||
| b | b2 | a | a2 |
| a2 | b2 | a | b | |||||
=( | −2+ | )+( | −2+ | )= | ||||
| b2 | a2 | b | a |
| a | b | |||
=( | − | )2+(√ab−√ba)2≥0 dla każdego a>0 i b>0⇔ | ||
| b | a |
| a | b | |||
Prawdziwa jest nierówność : | + | ≥2 ( będziemy z niej korzystać | ||
| b | a |
| a | a | b | b | ||||
+( | )2+ | +( | )2 | ||||
| b | b | a | a |
| a | b | a | b | a | b | |||||||
+ | +( | + | )2−2 | * | = | |||||||
| b | a | b | a | b | a |
| a | b | a | b | |||||
= | + | +( | + | )2− 2 ≥ 2+22−2=4 | ||||
| b | a | b | a |
| a | a2 | b | b2 | |||||
L = | + | + | + | = | ||||
| b | b2 | a | a2 |
| a | b | a2 | b2 | |||||
( | + | ) + ( | + | ) ≥ 2+2 = 4 | ||||
| b | a | b2 | a2 |
| 1 | ||
x+ | ≥ 2 | |
| x |
