Ratio of segments formed by intersection of three cevians

Question

ABC is a triangle and P some point inside it. Cevians AP, BP, CP cross the opposite sides at A',B' and C' respectively.

What is the minimum value the following expression can accomplish:

(1+2*PA'/PA)(1+2*PB'/PB)(1+2*PC'/PC) ?

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The solution is 8, when each ratio is 1/2 (meaning that is a centroid) but I'm having trouble reaching that.

Now from the ratio of areas APB, BPC, CPA and the whole triangle ABC we get respectively:

Pa/P = A'P/A'A, Pb/P = B'P/B'B, Pa/P = C'P/C'C, and summing this i get

1 = PA'/(PA' + PA) + PB'/(PB' + PB) + PC'/(PC' + PC)

Does from here directly follows some relation to the wanted expression fully algebraically, and if so how? Or should I still consider them as being in the triangle?

Replacing PA'/PA=x PB'/PB=y PC'/PC=z I get

1 = 1/(1 + 1/x) + 1/(1 + 1/y) + 1/(1 + 1/z), and the wanted expression as (1+2x)(1+2y)(1+2z)

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I also tried:

4 = 1 + x/(1+x) + 1 + y/(1+y) + 1 + z/(1+z)

4 = (1+2x)/(1+x) + (1+2y)/(1+y) + (1+2z)/(1+z)

Answer 1

Use the substitution $ a = \frac{1}{ 1 + \frac{1}{x} }$, or that $ x = \frac{a}{1-a}$, and $ 1 + 2x = \frac{ 1+a}{1-a}$.

Rephrasing what you already have

WTS $ \prod \frac{ 1 + a } { 1 - a } \geq 8 $ subject to $ a + b + c = 1$.

This is much easier to work with, give it a try.

Jensen's on the product: Show that $ \ln \frac{ 1+a}{1-a}$ is convex, so $\prod \frac{1+a}{1-a} \geq ( \frac{ 1 + 1/3 } { 1 - 1/3} ) ^3 = 8 $.

AM-GM: substituting in $ a+b+c = 1$, $\prod \frac{ (a+b) + (c+a) } { (b+c)} \geq \prod \frac{ 2 \sqrt{ (a+b)(c+a) }} { (b+c) } = 8.$

Answer 2

Starting from the following from the question I managed to push to the solution.

Replacing PA'/PA=x PB'/PB=y PC'/PC=z I get

1 = 1/(1 + 1/x) + 1/(1 + 1/y) + 1/(1 + 1/z) [1], and the wanted expression as

(1+2x)(1+2y)(1+2z)

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[1] can be written as 1 = xy + yz + xz + 2xyz. [3]

Also from [1] and H-A 1 <= (x+1)/4 + (y+1)/4 + (z+1)/4 that is 1 <= x + y + z [4]

Now using [3] and [4] we have

(1+2x)(1+2y)(1+2z) = 1 + 2x + 2y + 2z + 4xy + 4xz + 4yz + 8xyz = 1 + 2x + 2y + 2z + 4 >= 7

I did it on the paper, but then found mistake and it dropped from 8 to 7 :D hahaha