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.

[@ranshen1486]

(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.

[@kristijankocbek9080]

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.

[@jinhuiliao1137]

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

[@rachitbhalani9408]

Cauchy flipping makes it trivial

[@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

[@economicsresearch5666]

a natural explanation.

[@frozenicetea3494]

I dont think you can lower the denominator…

[@IITB-rt1xz]

great

[@kristijankocbek9080]

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.

[@jwy4264]

there are many ways to solve this problem

[@kristijankocbek9080]

@@jwy4264 So?

[@jwy4264]

@@kristijankocbek9080 well basically youre right, he did not have to do the +3 -3 trick, though, 3 is a handy number here because you know that that sum was equal to 3

[@mimo5313]

@@jwy4264 this is one of the ways

[@FaranAiki]

Because there are 3 variables which adds up to 1. If their sum were 2, add 6, so on.

[@imonkalyanbarua]

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

[@ekasurya8646]

Nice solution, but i think CS unnecessary.

[@wise_math]

@EkaSurya using rearrangement?

[@hsingh-13]

since a+b+c = 3
a

[@sovanraksa2112]

Your division inequality is not stable. a is indeed < 3 and b^2 + 1 is indeed < 10. That does not imply that a divided by b^2 + 1 is always less than 3/10. We can see this clearly by taking a = 2.9999 and b = 0.00001

[@varadthube4851]

try by am- hm inequality you will get it....

[@tiktokclips9704]

You're wrong in the part where you divided by b²+1 because b²+1 becomes >1/10 so you cannot combine between a1/10

[@intellectualbrainstorm665]

The question is indeed wrong. You have to prove that the LHS is lesser than or equal to the RHS 😊