If $\sum ab=1$, find the minimum of $\sum\sqrt{\frac a{8ab+1}}$

Question

Let $a$, $b$, $c$ be non-negative real numbers such that $ab+bc+ca=1$. Find the minimum of $$\sqrt{\frac a{8ab+1}}+\sqrt{\frac b{8bc+1}}+\sqrt{\frac c{8ca+1}}.$$ Source: Crux Mathematicorum

I tried the cases $a=b=c$ and $a=0$, and I guess that the minimum is $\frac2{\sqrt3}$, obtained when $(a,b,c)=\left(0,3,\frac13\right)$.

I thought of using Hölder inequality, but this might be a bad idea: If we apply something like $$\left(\sum_{\rm cyc}\sqrt{\frac a{8ab+1}}\right)^2\sum a^2(8ab+1)s_a^3\ge\left(\sum_{\rm cyc}as_a\right)^3,$$ then it is difficult to homogenize because the given condition is second degree.

Also I tried rearrangement inequality. Let $x\ge y\ge z$ be a permutation of $a$, $b$, $c$, then $xy+yz+zx=1$, and $$\sum_{\rm cyc}\sqrt{\frac a{8ab+1}}\ge\sqrt{\frac x{8xy+1}}+\sqrt{\frac y{8xz+1}}+\sqrt{\frac z{8yz+1}}.$$ Now it is half symmetric with $x$ and $z$. But the usual method $x+z=p$, $xz=q$ doesn't seem to help because the condition would turn into a complicated form $pc+q=1$.

Answer 1

By AM-GM $$\sum_{cyc}\sqrt{\frac{a}{8ab+1}}=\sqrt{\sum_{cyc}\frac{a}{8ab+1}+2\sum_{cyc}\sqrt{\frac{ab}{(8ab+1)(8bc+1}}}=$$ $$=\sqrt{\sum_{cyc}\frac{a}{8ab+1}+2\sqrt{\sum_{cyc}\frac{ab}{(8ab+1)(8bc+1}+2\sum_{cyc}\sqrt{\frac{ab^2c}{(8ab+1)(8bc+1)^2(8ca+1)}}}}\geq$$ $$\geq\sqrt{\sum_{cyc}\frac{a}{8ab+1}+2\sqrt{\sum_{cyc}\frac{ab}{(8ab+1)(8bc+1)}+6\sqrt[3]{\frac{a^2b^2c^2}{\prod\limits_{cyc}(8ab+1)^2}}}}$$ and it's enough to prove that $$\sum_{cyc}\frac{a}{8ab+1}+2\sqrt{\sum_{cyc}\frac{ab}{(8ab+1)(8bc+1)}+6\sqrt[3]{\frac{a^2b^2c^2}{\prod\limits_{cyc}(8ab+1)^2}}}\geq\frac{4}{3}$$ or $$\sum_{cyc}a(8bc+1)(8ca+1)+$$ $$+2\sqrt{\prod_{cyc}(8ab+1)\left(\sum_{cyc}ab(8ca+1)+6\sqrt[3]{a^2b^2c^2\prod_{cyc}(8ab+1)}\right)}\geq\frac{4}{3}\prod_{cyc}(8ab+1)$$

and since by $uvw$ $$\prod_{cyc}(8ab+1)\left(\sum_{cyc}ab(8ca+1)+6\sqrt[3]{a^2b^2c^2\prod_{cyc}(8ab+1)}\right)\geq\left(3+92abc+64a^2b^2c^2\right)^2,$$ it's enough t prove $$\sum_{cyc}a(8bc+1)(8ca+1)+6+184abc+128a^2b^2c^2\geq\frac{4}{3}\prod_{cyc}(8ab+1)$$ or $$3(a+b+c)-18+24\sum_{cyc}a^2c+816abc-256abc(a+b+c)\geq1664a^2b^2c^2$$ or $$\sum_{cyc}(3a^2b+27a^2c+275abc)\geq$$ $$\geq18(ab+ac+bc)^3+256abc(a+b+c)(ab+ac+bc)+1664a^2b^2c^2$$ or $$9(ab+ac+bc)^3(a^2b+b^2c+c^2a+9(a^2c+b^2a+c^2b)+275abc)^2\geq$$ $$\geq\left(18(ab+ac+bc)^3+256abc(a+b+c)(ab+ac+bc)+1664a^2b^2c^2\right)^2,$$ which is true by BW

Answer 2

Here is the solution done with Mathematica 14.2.1. Firstly, we switch to polynomials in the target function by the change $p^2=\frac a{8ab+1},r^2=\frac b{8bc+1},s^2=\frac c{8ca+1},p\ge0,r\ge 0,s\ge 0$. Then the target function becomes $p+r+s$. Secondly, we find where the restrictions hold by the MMA command (see the documentation for info)

CylindricalDecomposition[{(8*a*b + 1)*p^2 == a && (8*b*c + 1)*r^2 == 
b && (8*c*a + 1)*s^2 == c && a*b + b*c + c*a == 1 && a >= 0 && 
b >= 0 && c >= 0 && p >= 0 && r >= 0 && s >= 0}, {a, b, c, p, r, s}]

(a == 0 && b > 0 && c == 1/b && p == 0 && r == Sqrt[b/(1 + 8 b c)] && s == Sqrt[c]) || (a > 0 && ((0 <= b < 1/a && c == (1 - a b)/(a + b) && p == Sqrt[a/(1 + 8 a b)] && r == Sqrt[b/(1 + 8 b c)] && s == Sqrt[c/(1 + 8 a c)]) || (b == 1/a && c == (1 - a b)/(a + b) && p == Sqrt[a/(1 + 8 a b)] && r == Sqrt[b/(1 + 8 b c)] && s == -Sqrt[(c/(1 + 8 a c))]))) $$\left(a=0\land b>0\land c=\frac{1}{b}\land p=0\land r=\sqrt{\frac{b}{8 b c+1}}\land s=\sqrt{c}\right)\lor \left(a>0\land \left(\left(0\leq b<\frac{1}{a}\land c=\frac{1-a b}{a+b}\land p=\sqrt{\frac{a}{8 a b+1}}\land r=\sqrt{\frac{b}{8 b c+1}}\land s=\sqrt{\frac{c}{8 a c+1}}\right)\lor \left(b=\frac{1}{a}\land c=\frac{1-a b}{a+b}\land p=\sqrt{\frac{a}{8 a b+1}}\land r=\sqrt{\frac{b}{8 b c+1}}\land s=-\sqrt{\frac{c}{8 a c+1}}\right)\right)\right) $$

We see that the restrictions split in two ones.

Thirdly, we consider the first case

Minimize[{p + r + s, (a == 0 && b > 0 && c == 1/b && p == 0 && 
r == Sqrt[b/(1 + 8 b c)] && s == Sqrt[c])}, {a, b, c, p, r, s}]

{2/Sqrt[3], {a -> 0, b -> 3, c -> 1/3, p -> 0, r -> 1/Sqrt[3], s -> 1/Sqrt[3]}} $$\left\{\frac{2}{\sqrt{3}},\left\{a\to 0,b\to 3,c\to \frac{1}{3},p\to 0,r\to \frac{1}{\sqrt{3}},s\to \frac{1}{\sqrt{3}}\right\}\right\} $$

Fourthly, we consider the second case by

NMinimize[{p + r +  s, (a > 0 && ((0 <= b < 1/a && c == (1 - a b)/(a + b) && 
    p == Sqrt[a/(1 + 8 a b)] && r == Sqrt[b/(1 + 8 b c)] && 
    s == Sqrt[c/(1 + 8 a c)]) || (b == 1/a && 
    c == (1 - a b)/(a + b) && p == Sqrt[a/(1 + 8 a b)] && 
    r == Sqrt[b/(1 + 8 b c)] && 
    s == -Sqrt[(c/(1 + 8 a c))])))}, {a, b, c, p, r, s}]

{1.19043, {a -> 0.57735, b -> 0.57735, c -> 0.57735, p -> 0.396811, r -> 0.396811, s -> 0.396811}}

In the above we use the numeric command NMinimize since Minimize is running without any response for a long time. Because $\frac{2}{\sqrt{3}} < 1.19043$ we can draw the conclusion that one of 6 optimal solutions is $a=0,, b=3, c=\frac 1 3$. I think all that can be mimicked by hand.

Addition. The doubts of @RiverLi concerning a local minimum instead of a global one produced by

NMinimize[{p + r +  s, (a > 0 && ((0 <= b < 1/a && c == (1 - a b)/(a + b) && 
p == Sqrt[a/(1 + 8 a b)] && r == Sqrt[b/(1 + 8 b c)] && 
s == Sqrt[c/(1 + 8 a c)]) || (b == 1/a && 
c == (1 - a b)/(a + b) && p == Sqrt[a/(1 + 8 a b)] && 
r == Sqrt[b/(1 + 8 b c)] && 
s == -Sqrt[(c/(1 + 8 a c))])))}, {a, b, c, p, r, s}]

came true. Indeed, the result of the command

NMinimize[{p + r + s, a > 0 && r >= 0 && p >= 0 && 
  s >= 0 && (0 <= b < 1/a && c == (1 - a b)/(a + b) && 
 p^2 == a/(1 + 8 a b) && r^2 == b/(1 + 8 b c) && 
 s^2 == c/(1 + 8 a c))}, {a, b, c, p, r, s}]

{1.1547, {a -> 0.333334, b -> 6.04174*10^-12, c -> 2.99999, p -> 0.577351, r -> 2.66924*10^-9, s -> 0.57735}}

(which leads to the exact solution $a= \frac 1 3, b=0, c=3$) shows the minimum in the second case is not greater than the one in the first case. A happy end is achieved through the Mathematica command FindInstance(see its documentation), where we exclude c and return to the target function with square roots,

FindInstance[a > 0 && 0 <= b < 1/a &&    2/Sqrt[3] > Sqrt[a/(1 + 8 a b)] + Sqrt[b/(1 + 8 b c)] + 
 Sqrt[c/(1 + 8 a c)] /. c -> (1 - a b)/(a + b), {a, b}, Reals]

{}

i.e. the value of the target function less than $\frac 2 {\sqrt{3}}$ cannot be obtained.