Question

A,b,c are distunct positive no. Such that a+b+c=1 then least value of (1+a)(1+b)(1+c)/ (1-a)(1-b)(1-c) is

Answer 1

Answer:

Answer:

1

Step-by-step explanation:Your Adding and then dividing by the same number which will make the distinct positive 1.

Step-by-step explanation:

Answer 2

Answer:

8

This value occurs when a = b = c = 1/3.

Step-by-step explanation:

By substitution of a = b = c = 1/3, it is clear that the value of 8 is attained.  We only need to prove then that

(1+a)(1+b)(1+c) / (1-a)(1-b)(1-c)  ≥ 8

Or equivalently, that

(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)

[ we can multiply both sides as we have since 1-a, 1-b, 1-c > 0 ].

Put x = 1+a, y = 1+b, z = 1+c.

Then x+y = 2+a+b = > 2 > 1+c = z

and similarly

y+z > x and z+x > y.

Therefore x, y, z are the sides of a triangle.

Let s = (x+y+z)/2 be the semiperimeter of the triangle, r the inradius, R the circumradius and Δ the area.

Then

2(1-a) = 2-2a = 1+a+b+c-2a = 1-a+b+c = -a-1+b+1+c+1 = -x+y+z = (x+y+z)-2x = 2(s-x)  =>  1-a = s-x

Similarly 1-b = s-y and 1-c = s-z.

So the inequality that we need to prove becomes

(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)

<=> xyz ≥ 8(s-x)(s-y)(s-z)

Some useful formulas for the area of a triangle include

• Δ² = s(s-x)(s-y)(s-z)      [ that's Heron's Formula ]
• Δ = xyz / 4R
• Δ = rs

Also, we should know that

• R ≥ 2r, with equality if and only if the triangle is equilateral.  [ One way to obtain this is use the fact that the distance d between incentre and circumcentre satisfies d² = R² - 2Rr, which is then positive. ]

Using these, we have

R ≥ 2r

=> Rs ≥ 2rs = 2Δ

=> 4RΔs ≥ 8Δ²

=> xyzs ≥ 8s(s-x)(s-y)(s-z)

=> xyz ≥ 8(s-x)(s-y)(s-z)

which is what we needed to prove.  Furthermore, we have equality if and only if the triangle is equilateral, so if and only if x = y = z, so if and only if a = b = c = 1/3.