Mayhem Solutions: M146-M150
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a+b+c = 1
1
4
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√
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√
√ ≤
ab + bc + ca (a2 b2 + b2 c2 + c2 a2 )
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√
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√
bc + ca ≤
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1
(a
2
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1
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≤
2
2
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1
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≥ c2 (a + b)2 = a2 c2 + b2 c2 + 2abc2 > a2 c2 + b2 c2 + abc2
a ≤ b ≤ c Z/
X%\^i1e^X
abc2 > a2 b2 Z
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r
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= πr 2 1 −
Z
n
n
n
$
2
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r
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nπ
= πr
n
n
1
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Z n=4
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n
3 =
1
2
πr
n
1
= n 1−
= n−1
n
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x > 1Z y > 1Z z > 1
x = yz
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logzx xy 4 z
logxy xyz 4
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g = logxy (xyz 4 )
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q
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g
=
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a = 2xyt Z
b = (x2 − y 2 )t Z
c = (x2 + y 2 )t Z
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A
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2
2
2
2
c = 2xyt + (x − y )t + (x + y$ )ta = Y+2xt(x
i/f P =1 a + b +
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h
A = 2 ab = 12 (2xyt)(x2 − y 2 )t = xyt2 (x2 − y 2 )
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λ
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xyt2 (x2 − y 2 ) = λ 2xt(x + y)
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t
y
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y(x − y)t
2
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z1
z1 + z 2
z1 · z 2
i/f
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z2 Z
f
i/f
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z2
= −(i$ + 1) Z
= −i
z1 · z 2
$
$
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X
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z
2
2
z1 · z2 = 1 · |z2 |
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z
2
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z1 + 4z1 z2 + z22 = 0 Z
Z
z1
z2
2
+4
z1
z2
+1 = 0
$
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Z z1 · z 2
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|z1 · z2 | = |z1 · z2 | = 1 Z/
z1 · z2 = −1