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what is the minimum value of a/(b+c)+b/(c+a)+c/(a+b)?
1 Answer
- IndicaLv 73 years ago
Global minimum for positive reals only.
Cauchy : (a/(b+c)+b/(a+c)+c/(a+b)) * (a(b+c)+b(a+c)+c(a+b)) ≥ (a+b+c)²
∴ (a/(b+c)+b/(a+c)+c/(a+b)) ≥ ½(a+b+c)²/(ab+bc+ca) … (i)
But because a²+b²+c² ≥ ab+bc+ca so (a+b+c)² ≥ 3(ab+bc+ca) and (a+b+c)²/(ab+bc+ca) ≥ 3
RHS of (i) = 3/2 and equality is achieved when a=b=c so minimum = 3/2
Another way is to assume WLOG that a≤b≤c and use Rearrangement Inequality
a≤b≤c ⟹ 1/(b+c) ≤ 1/(a+c) ≤ 1/(a+b) so applying it twice
a/(b+c)+b/(a+c)+c/(a+b) ≥ c/(b+c)+a/(a+c)+b/(a+b)
a/(b+c)+b/(a+c)+c/(a+b) ≥ b/(b+c)+c/(a+c)+a/(a+b)
Sum these for 2{a/(b+c)+b/(a+c)+c/(a+b)} ≥ 3
