Find all real numbers r, such that the inequality r(ab+bc+ca)+(3−r)(1a+1b+1c)≥9holds for any real a,b,c>0.
Problem
Source: 2007 Bulgarian Autumn Math Competition, Problem 12.3
Tags: Symmetric inequality, three variable inequality, algebra, inequalities
19.03.2022 21:06
Bump to this beauty
19.03.2022 21:47
I claim the only such r is r=1. Step 1: r<0 and r>3 are impossible. Note that if r<0 then by making a,b,c arbitrarily large, we obtain a contradiction. Likewise, if r>3, then we obtain a similar contradiction by making a,b,c arbitrarily small. Thus 0≤r≤3. Step 2. Impossibility of 0≤r≤3 except r=1. Now, see that r=0,3 does not work. Assume r∈(0,3). Take a,b,c such that a3=b3=c3=3−r2r.With this, r(ab+bc+ca)+(3−r)(1a+1b+1c)=(ab+bc+ca)(r+3−ra3)=9r(3−r2r)23.Now, we show this quantity is always below 9. Assume the contrary and obtain r3(3−r)24r2≥1⟺r(3−r)2≥4⟺(r−4)(r−1)2≥0,implying r≥4 or r=1. But we already shown r>3 is impossible. Hence, the only remaining case is r=1. We claim r=1 works. To that end, it suffices to show (ab+bc+ca)(1+2abc)≥9.Set abc=t3 and note by AM-GM that ab+bc+ca≥3t2. With this, it suffices to have 3t2(1+2t3)≥9⟺(t−1)2(t+2)t≥0,which is trivially true as t>0.
19.03.2022 22:26
20.03.2022 04:21
Marinchoo wrote: Find all real numbers r, such that the inequality r(ab+bc+ca)+(3−r)(1a+1b+1c)≥9holds for any real a,b,c>0. https://artofproblemsolving.com/community/c6h476867p24717131