More about completing square

1)x⁴+ 6x² - 7
= x⁴+ 2(3)x² + 9 - 16
= (x² + 3)² - 16
≥ (0 + 3)² - 16
= - 7
∴ The minimum value of x⁴+ 6x² - 7 is 7.
Alternatively :
x⁴+ 6x² - 7
= (x²)² + 6x² - 7 ≥ 0² + 6(0)² - 7 = - 7 .

2)...... 2x
───────
x² + x + 1........... 2
= ───────
... x + 1 + 1/x
............. .......2
= ──────────────
... x + 2√x(1/√x) + 1/x - 1
.............. 2
= ──────────
... (√x + 1/√x)² - 1
When (√x + 1/√x)² = 0 , f(x) attains its minimum.
i.e.
x + 1/x + 2 = 0
x² + 2x + 1 = 0
(x + 1)² = 0
x = - 1∴ f(x) attains its minimum at x = - 1.
...... 2x
───────
x² + x + 1........... 2
= ───────
... x + 1 + 1/x
............. .......2
= ──────────────
... x - 2√x(1/√x) + 1/x + 3
............. ..2
= ──────────
... (√x - 1/√x)² + 3
When (√x - 1/√x)² = 0 , f(x) attains its maximum.
i.e.
x + 1/x - 2 = 0
x² - 2x + 1 = 0
(x - 1)² = 0
x = 1 ∴ f(x) attains its maximum at x = 1.
3)√x + √(y - 1) + √(z - 2) = (x + y + z) / 2
x - 2√x + y - 2√(y - 1) + z - 2√(z - 2) = 0
(x - 2√x + 1) + (y - 1 - 2√(y - 1) + 1) + (z - 2 - 2√(z - 2) + 1) = 1 - 1 + 1 - 2 + 1
(√x - 1)² + (√(y - 1) - 1)² + (√(z - 2) - 1)² = 0
⇒
(√x - 1)² = (√(y - 1) - 1)² = (√(z - 2) - 1)² = 0
√x = 1 and √(y - 1) = 1 and √(z - 2) = 1
x = 1 and y = 2 and z = 3

The first two questions are answered here:
http://imgur.com/qq0Wcro

For the last one, not much idea.

Seems that the level is different from the other two?

2013-10-17 00:59:28 補充：
我又慢左啦~
只可以排第三......

2013-10-17 01:07:06 補充：
好多謝你的指導呀~

http://imgur.com/rlrr7hP