a nice but little tricky inequality (math olympiad problem)

This is a math olympiad inequality problem I found in a training for Math Olympiad teachers, in which I did the grading for the test. If anyone found where did this math olympiad problem appear, please let me know: the contest, or book, etc.

This inequality is quite tricky, but once you find a way, the proof is actually simple and quite short. A crucial step is the simple algebraic manipulation. Then using Cauchy Schwarz or could also Rearrangement in the middle of proof.

I make videos about math olympiad or contest mathematics problems, but not limited to it I also make videos on calculus, pure, and applied mathematics.

[ranshen]

(1) Alternative to the first step: Since f(x)=1/(x²+1) is convex, we may consider using Jensen's inequality.
Define weights p₁=a/3, p₂=b/3, & p₃=c/3, and thus their sum is 1. Define locations x₁=b, x₂=c, & x₃=a. Thus,
LHS/3 = p₁f(x₁) + p₂f(x₂) + p₃f(x₃)
≥ f( p₁x₁ + p₂x₂ + p₃x₃ )
= f{ (ab+bc+ac)/3 }
= 1 / [ {(ab+bc+ac)/3}² +1 ].
It remains to show ab+bc+ac ≤ 3.

(2) Alternative to the 2nd step: ab+bc+ac and a+b+c suggest using Vieta's formulas for the cubic equation. Define q=ab+bc+ac. Then a, b, & c are real roots of
x³ - 3x² + qx -abc =0.
We might check the cubic discriminant but that's messy and unnecessary for this problem. To have 3 real roots, it is nececcary (but not sufficient) for the cubic to have (1 or 2) stationary points. We thus differentiate the cubic first to get
3x² - 6x + q =0.
Its (quadratic) discriminant is 6² - 4×3q = 4(9 - 3q). To have stationary points, this quadratic discriminant needs to be ≥ 0, i.e., q ≤ 3.

[kristijankocbek]

Another more elementary solution is using tangent line method: 1/(1+x^2)=> 1-x/2 Now use well know a^2+b^2+c^2\geq ab+bc+ca.

[jinhuiliao]

a/(b^2+1)+b/(c^2+1)+c/(a^2+1)
>=a/2b+b/2c+c/2a (AM-GM)
Without losing generality, let a<=b<=c, 1/2c<=1/2b<=1/2a
(Rearrangement inequality Reverse ordered)
=3/2
The equality holds when b^2=1, a^2=1, c^2=1.

[frozenicetea]

I dont think you can lower the denominator…

[OCEAN_NINJA]

I tried myself it for half hour, when
after a lot thinking I am not getting the sol. pressed space bar he said :we might to recall this famous and very simple inequality" my hart(heart) broke down

[imonkalyanbarua]

I am sorry but I didn't get clarity with this solution. 🙏

[ekasurya]

Nice solution, but i think CS unnecessary.

[hsingh-]

since a+b+c = 3
a<3
b<3 b²+1<10
c<3

Therefore,
a ÷ (b²+1) < 3/10
Similarly,
b ÷ (c²+1) < 3/10
c ÷ (a²+1) < 3/10

By adding all three,

{a ÷ (b²+1)} + {b ÷ (c²+1)} + {c ÷ (a²+1)} < 3 ×3/10 = 9/10
Hence, the sum is less than 9/10 i.e. even less than 1 but it's asked to prove it greater than equal to 3/2.

Therefore, the question is wrong.

[kristijankocbek]

This solution is completly unnatural. How could someone think of adding 3-3? Why not 1-1? There must be more instructive way than this divine intervention.