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Junior Balkan Mathematical Olympiad

Algebra Difficulty 5.0 AIME Prove it JBMO

Problem:
Find all triples of positive real numbers (a,b,c)(a, b, c) so that the expression
M=(a+b)(b+c)(a+b+c)abc M=\frac{(a+b)(b+c)(a+b+c)}{a b c}
gets its least value.

Solution

Solution:
The expression MM is homogeneous, therefore we can assume that abc=1a b c=1. We set s=a+cs=a+c and p=acp=a c and using b=1acb=\frac{1}{a c}, we get
M=(a+1ac)(1ac+c)(a+1ac+c)=(a+p−1)(c+p−1)(s+p−1) M=\left(a+\frac{1}{a c}\right)\left(\frac{1}{a c}+c\right)\left(a+\frac{1}{a c}+c\right)=\left(a+p^{-1}\right)\left(c+p^{-1}\right)\left(s+p^{-1}\right)
Expanding the right-hand side we get
M=ps+s2p+1+2sp2+1p3 M=p s+\frac{s^{2}}{p}+1+\frac{2 s}{p^{2}}+\frac{1}{p^{3}}
Now by s≥2ps \geq 2 \sqrt{p} and setting x=pp>0x=p \sqrt{p}>0 we get
M≥2x+5+4x+1x2 M \geq 2 x+5+\frac{4}{x}+\frac{1}{x^{2}}
We will now prove that
2x+5+4x+1x2≥11+552. 2 x+5+\frac{4}{x}+\frac{1}{x^{2}} \geq \frac{11+5 \sqrt{5}}{2} \text{.}
Indeed, the latter is equivalent to 4x3−(55+1)x2+8x+2≥04 x^{3}-(5 \sqrt{5}+1) x^{2}+8 x+2 \geq 0, which can be rewritten as
(x−1+52)2(4x+3−5)≥0 \left(x-\frac{1+\sqrt{5}}{2}\right)^{2}(4 x+3-\sqrt{5}) \geq 0
which is true.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.