AM/GM Inequality!!!?

Question

)If a, b, c are all real numbers, all lying between 0 and 1, such that a+b+c = 2, show that

abc
________>=8
(1-a)(1-b)(1-c)

Answer 1

By AM/GM Inequality,

a+b+c >= (3/```abc``)x3 (R.H.S. is thrice the cube root of abc )
Since a+b+c =2,
2 >= (3/```abc``)x3
By cubing both sides,
8 >= abc/27
-abc/27 >= 8 .....(1)

Also, (1-a)+(1-b)+(1-c) >= (3/```(1-a)(1-b)(1-c)``)x3
:. 3-a-b-c >= 27(1-a)(1-b)(1-c)
:. 1>= 27(1-a)(1-b)(1-c).......(2)

From1 and 2, hence proved

Answer 2

Let a = b = c
a + b + c = 2
so 3 a = 2 or a = 2/3
a × b × c = a^3 = (2/3)^3 = 8/27
(1 – a)(1 – b)(1 – c) = (1 – a)(1 – a)(1 – a) = (1 – a)^3
= (1 – 2/3)^3 = (1/3)^3 = 1/27
a b c / (1 – a)(1 – b)(1 – c) = (8/27) / (1/27) = 8
The minimum value of a = 0..3 then b = 0.9 and c = 0.8
a × b × c = 0..3 × 0.9 × 0.8 = 0.216
(1 – a)(1 – b)(1 – c) = (1 – 0.3)(1 – 0.9)(1 – 0.8)
= 0.7 × 0.1 × 0.2 = 0.014
a b c / (1 – a)(1 – b)(1 – c) = 0.216 / 0.014 = =15.4
Show that the result is = 8 or greater than 8

Answer 3

he AM-GM Inequality
For positive numbers, the arithmetic mean is greater than or equal to the geometric mean

Proof using Jensen's Inequality:

Jensen's Inequality is

∑(f(xi))/n ≤ f(∑(xi)/n) if f is concave (i.e. f''(x)<0, which is sometimes called "concave down"),

∑(f(xi))/n ≥ f(∑(xi)/n) if f is convex (i.e. f''(x)>0, which is sometimes called "concave up"),

with equality iff x1 = x2 = ... = xn.

ln(x) is concave, so by Jensen's Inequality,

(ln(a1) + ln(a2) + ... + ln(an))/n ≤ ln((a1+a2+...+an)/n),

Taking the exponential function of both sides,

geometric mean(a1,a2,...,an) ≤ arithmetic mean(a1,a2,...,an)

Weighted AM-GM Inequality

Given n positive reals a1,a2,...,an, and n weights w1,w2,...,wn such that 0≤wi≤1 and ∑wi=1,
w1a1+w2a2+...+wnan ≥ a1w1 a2w2 ... anwn

Proof Outline: same as above, using the weighted version of Jensen's Inequality; or use the ordinary AM-GM inequality for rational weights (put all the weights over a common denominator, d, and then show that the AM of the d numbers (many identical) is never less than the GM of those same numbers); and then extend to irrational weights by continuity.
Internet References

Mathworld -- Jensen's Inequality

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Answer 4

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Answer 5

a=0.5, b=0.7,c=0.8

abc=0.28
(1-a)(1-b)(1-c)=0.12