Let a,b, and c be positive real numbers such that ab+bc+ca=3. Show that bc1+a4+ca1+b4+ab1+c4≥32.
Problem
Source: Philippine MO 2022/7
Tags: inequalities, algebra
19.03.2022 08:37
I can't believe I had to do this again. Claim: 11+c2≥1−c2 for c≥0. Proof: Just expand, and get 2≥(2−c)(1+c2)⟺c(c−1)2≥0. Equality when c=0,1. Hence, 11+x4≥2−x22 with equality at x=1 for x>0. Write∑cycbc1+a4≥∑(bc−a2bc2)=3−12abc(a+b+c)so it suffices to showabc(a+b+c)≤3⟺3abc(a+b+c)≤(ab+bc+ca)2=a2b2+b2c2+c2a2+2abc(a+b+c)which reduces toabc(a+b+c)≤a2b2+b2c2+c2a2which is simply just the known inequality xy+yz+zx≤x2+y2+z2 for x=ab,y=bc,z=ca, so we are done.
19.03.2022 08:43
not needed ..
19.03.2022 09:23
bc=x,ab=z,ac=ybc1+a4=x3x2+y2z2=x−xy2z2x2+y2z2 ∑xy2z2x2+y2z2≤∑xy2z22xyz≤32◼
19.03.2022 15:06
Let a,b, and c be positive real numbers such that ab+bc+ca=3. Show thatbc1+a2+ca1+b2+ab1+c2≥32
20.03.2022 09:40
sqing wrote: Let a,b, and c be positive real numbers such that ab+bc+ca=3. Show thatbc1+a2+ca1+b2+ab1+c2≥32
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20.03.2022 09:52
jj_ca888 wrote: I can't believe I had to do this again. Claim: 11+c2≥1−c2 for c≥0. Proof: Just expand, and get 2≥(2−c)(1+c2)⟺c(c−1)2≥0. Equality when c=0,1. Hence, 11+x4≥2−x22 with equality at x=1 for x>0. Write∑cycbc1+a4≥∑(bc−a2bc2)=3−12abc(a+b+c)so it suffices to showabc(a+b+c)≤3⟺3abc(a+b+c)≤(ab+bc+ca)2=a2b2+b2c2+c2a2+2abc(a+b+c)which reduces toabc(a+b+c)≤a2b2+b2c2+c2a2which is simply just the known inequality xy+yz+zx≤x2+y2+z2 for x=ab,y=bc,z=ca, so we are done. Solution indeed. Bravo
20.03.2022 15:53
InternetPerson10 wrote: Let a,b, and c be positive real numbers such that ab+bc+ca=3. Show that bc1+a4+ca1+b4+ab1+c4≥32.
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21.03.2022 11:34
Let x=bc,y=ca,z=ab, then x+y+z=3. Then ∑cycbc1+a4=∑cycx1+(yz/x)2=∑cycx3x2+y2z2=∑cycx−xy2z2x2+y2z2=3−∑cycxy2z2x2+y2z2≥3−∑cycxy2z22xyz=3−12(xy+yz+zx)≥3−12⋅(x+y+z)23=32.
30.03.2022 05:47
Here is a brute forced solution XD (took less time than i thought) Using Cauchy we have that: bc1+a4+ac1+b4+ab1+c4≥93+abc(a3+b3+c3)So now we need to prove that 93+abc(a3+b3+c3)≥32which is equivalent to3≥abc(a3+b3+c3)So now we use Lagrange multipliers, define g(a,b,c)=ab+bc+ac−3 and f(a,b,c)=a3+b3+c3 so now, for a constant λ∈R we define L(a,b,c,λ)=f(a,b,c)+λg(a,b,c)=a3+b3+c3+λ(ab+bc+ca−3) so now we need the gradient to be 0 which is equivalent to find a,b,c,λ so that ab+bc+ac=3 and 3a2=−λ(b+c)3b2=−λ(a+c)3c2=−λ(a+b)If a≠b≠c then by substtracting one of those equations with the other we end up with 3a+3c=3b+3c=3a+3b=λ which means a=b=c which is a contradiction so at least from the numbers a,b,c one is equal to the other one, lets say a=b WLOG so now if a≠c then by substracting 2 equations we get 3a+3c=λ but also in one of the equations we get λ=−3c22a but negative cant be equal to positive so we get a contradiction, meaning that a=b=c so a=b=c=1 so the minimun of f down condition g is f(1,1,1)=3 so 3≥a3+b3+c3 and now by Am-Gm we have 3=ab+bc+ac≥3(abc)23 which means 1≥abc so 3≥a3+b3+c3≥abc(a3+b3+c3)Thus we are done
30.03.2022 06:50
MathLuis wrote:
For Lagrange Multipliers, you have to show that a minimum of f exists.
22.01.2023 19:56
InternetPerson10 wrote: Let a,b, and c be positive real numbers such that ab+bc+ca=3. Show that bc1+a4+ca1+b4+ab1+c4≥32. We need to prove that ∑bca4+1−bc=−∑bca4a4+1≥−32⇔∑bca4a4+1≤32 with is true because: \sum \frac{bca^4}{a^4+1}\leqslant \sum \frac{bca^4}{2a^2}= abc(a=b+c)\leqslant \frac{(ab+bc+ca)^2}{3}=3