Let $a+b+c=0$ and $\{a,b,c\}\subset[-1,1]$. Prove that: $$\sqrt{1+a+\frac{7}{9}b^2}+\sqrt{1+b+\frac{7}{9}c^2}+\sqrt{1+c+\frac{7}{9}a^2}\geq3$$ I tried Holder and more, but without success.
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you can plug $$c=-a-b$$ in the left hand side and consider a problem in two variables – Dr. Sonnhard Graubner Dec 23 '16 at 19:36
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@Dr. Sonnhard Graubner I tried. It's nothing. – Michael Rozenberg Dec 23 '16 at 19:38
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i have found $$a=b=c=0$$ with my PC – Dr. Sonnhard Graubner Dec 23 '16 at 19:39
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or you use the Lagrange multiplier function – Dr. Sonnhard Graubner Dec 23 '16 at 19:40
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@Dr. Sonnhard Graubner I look for an human proof. – Michael Rozenberg Dec 23 '16 at 19:40
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i'm also looking for such a proof – Dr. Sonnhard Graubner Dec 23 '16 at 19:40
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@Michael Rozenberg: If $\sqrt{1+a+\frac{7}{9}b^2}\cdot\sqrt{1+b+\frac{7}{9}c^2}\cdot\sqrt{1+c+\frac{7}{9}a^2} = 1$, then the claim follows immediately. – Moritz Dec 23 '16 at 19:42
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@Moritz are you sure that $\prod\limits_{cyc}\left(1+a+\frac{7}{9}b^2\right)\geq1$ is true? – Michael Rozenberg Dec 23 '16 at 19:46
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@Michael Rozenberg: No, I am not sure, that the product in your example really has the value 1 but IF you can show it to be true (exactly $=1$), then you are finished. – Moritz Dec 23 '16 at 19:48
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@Moritz i think you should take a look first what other inequalities the OP proposed, how many of them were solved, and the solutions of the solved ones :) – Momo Dec 24 '16 at 03:40
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Where does this inequality come from? – Piquito Dec 24 '16 at 23:49
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@Piquito I think it was in AoPS. – Michael Rozenberg Dec 25 '16 at 02:40
1 Answers
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This one we can use IntervalRootFinding.jl, which "it guarantees to find all roots inside a given box".
The inequality is equivalent to $$ f(a,b):=\sqrt{\frac{7 a^2}{9}-a-b+1}+\sqrt{a+\frac{7 b^2}{9}+1}+\sqrt{\frac{7}{9} (-a-b)^2+b+1}-3 \ge 0. $$ This not defined on the whole $T=[-1,1]^2$, but that's fine.
It's easy to check that $f(a,b)$ has a critical point at $(0,0)$. Using the following code, we can find all critical points of $f(a,b)$ in $T$ and we see that there is no minimal point inside $T$.
using IntervalArithmetic, IntervalRootFinding, ForwardDiff
f((a,b))=-3 + sqrt(1 - a + (7*a^2)/9 - b) + sqrt(
1 + 7/9*(-a - b)^2 + b) + sqrt(1 + a + (7*b^2)/9)
∇f = ∇(f)
rts = roots(∇f, IntervalBox(-5..6, 2), Newton, 1e-5)
pt=interval.(rts)
println("The critical points")
println(pt)
println("The values at critical points")
println(f.(pt))
The output is
The critical points
IntervalBox{2,Float64}[[0.187395, 0.187396] × [0.665432, 0.665433], [0.665432, 0.665433] × [-0.852829, -0.852828], [-0.852829, -0.852828] × [0.187395, 0.187396], [-1.46504e-09, 1.46266e-09] × [-2.14219e-09, 2.13365e-09]]
The values at critical points
Interval{Float64}[[0.149065, 0.149067], [0.149065, 0.149066], [0.149065, 0.149066], [-3.60177e-09, 3.60177e-09]]
It remains to check the boundary of $T$, which is quite easy.
NonalcoholicBeer
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