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

Another way.

Lemma. Let $x$, $y$ and $z$ be non-negative numbers such that $$x^2y^2+x^2z^2+y^2z^2+10x^2y^2z^2\geq\frac{1}{9}.$$ Prove that $$x+y+z\geq\frac{2}{\sqrt3}.$$

By using this lemma it's enough to prove:

$$\sum_{cyc}\frac{ab}{(8ab+1)(8bc+1)}+\frac{10abc}{\prod\limits_{cyc}(8ab+1)}\geq\frac{1}{9}$$ for any non-negatives $a$, $b$ and $c$ such that $ab+ac+bc=1$.

A proof of the the lemma.

Let $x+y+z<\frac{2}{\sqrt3},$ $x=kp$, $y=kp$ and $z=kp$, where $k>0$ and $$p+q+r=\frac{2}{\sqrt3}.$$ Thus, $$k(p+q+r)<\frac{2}{\sqrt3},$$ which gives $0<k<1.$

But $$\frac{1}{9}\leq\sum_{cyc}x^2y^2+10x^2y^2z^2=k^4\sum_{cyc}p^2q^2+10k^6p^2q^2r^2<$$ $$<\sum_{cyc}p^2q^2+10p^2q^2r^2,$$ which is a contradiction because we'll prove now: $$\sum_{cyc}p^2q^2+10p^2q^2r^2\leq\frac{1}{9}$$ or $$\frac{3}{4}(p+q+r)^2\sum_{cyc}p^2q^2+10p^2q^2r^2\leq\frac{3}{64}(p+q+r)^6$$ or $$3(p+q+r)^6\geq48(p+q+r)^2\sum_{cyc}p^2q^2+640p^2q^2r^2,$$ which is in the uvw' notation $f(u)\geq0,$ where $$f(u)=2187u^6-48\cdot9\cdot3u^2(3v^4-2uw^3)-640w^6.$$ But by Maclaurin $$f'(u)=6\cdot2187u^5-48\cdot9\cdot18uv^4+48\cdot9\cdot18u^2w^3=$$ $$=486u(27u^4+16uw^3-16v^4)\geq0,$$ which says that $f$ increases.

Thus, by uvw it's enough to prove $f(u)\geq0$ for an equality case of two variables,

for which it's enough to assume $q=r=1$, which gives: $$p(3p^5+36p^4+84p^3+96p^2-352p+384)\geq0,$$ which is true because even $$176^2-96\cdot384<0,$$ which finishes a proof of the lemma.

Id est, it remains to prove $$9\sum_{cyc}ab(8ac+1)+90abc\geq\prod_{cyc}(8ab+1)$$ or $$9(ab+ac+bc)\sum_{cyc}ab(9ac+ab+bc)-\prod_{cyc}(9ab+ac+bc)+90abc\geq0$$ or $$4\left(\sum_{cyc}(a^2b+a^2c)-61abc\right)+45\geq0,$$ which is true by AM-GM: $$4\left(\sum_{cyc}(a^2b+a^2c)-61abc\right)+45=$$ $$=4\left(\sum_{cyc}(a^2b+a^2c)-61abc\right)+45\sqrt{(ab+ac+bc)^3}\geq$$ $$\geq(4\cdot(6-61)+45\sqrt{27})abc\geq0, $$ which finishes the solution.

Answer 3

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.

Answer 4

I tried some Calculus, but ended having to use Wolfram. At least now I am convinced the answer is as found: on the boundary! The long story:

Let $$ u := \sqrt{\frac{a}{8ab+1}}+\sqrt{\frac{b}{8bc+1}}+\sqrt{\frac{c}{8ca+1}}. $$ The question asks to find the minimum of $u$ subject to $$ C := \{ a,b,c \ge 0:\ ab + bc + ca = 1\}. $$ If some, say $c$, is zero, then $$ u(a,1/a,0) = \frac{\sqrt{a}}{3} + \frac 1{\sqrt{a}} \ge \frac{2}{\sqrt 3}, $$ attained at $a= 3$.

So, we can assume that $abc > 0$ and look for minimums smaller than $2/\sqrt 3$ in the interior of the first octant.

Note that $a+b>0$ on $C$, so we can substitute $$ c(a,b) := \frac{1-ab}{a+b} $$ to get $$ u(a,b,c(a,b)) = v(a,b) = \sqrt{\frac{a}{8ab+1}} + \sqrt{\frac{b(a+b)}{a+9b-8ab^2}} + \sqrt{\frac{1-ab}{9a+b-8a^2b}} $$ over the domain $$ D:\begin{cases} 0 < a < \infty\\ 0 < b < 1/a\\ \end{cases} $$ Substituting $$ x = a,\quad y = ab, $$ we get $$ v(a,b) = f(x,y) := \sqrt{\frac{x}{8y+1}} + \sqrt{\frac{y(x^2 + y)}{x(x^2 + 9y - 8y^2)}} + \sqrt{\frac{(1-y)x}{9x^2 + y - 8x^2y}} $$ over the rectangle $$ E:\begin{cases} 0 < x < \infty\\ 0 < y < 1\\ \end{cases} $$ The command Plot3D[Sqrt[x/(8*y + 1)] + Sqrt[y*(x^2 + y)/(x*(x^2 + 9*y - 8*y^2))] + Sqrt[(1 - y)*x/(9*x^2 + y - 8*x^2*y)], {x, 0.01, 1}, {y, 0.01, 0.5}] on www.wolframalpha.com revelas that there is sharp minimum at the bounday - the above - and a flat one inside with a saddle point (``mountain pass'') in between. So, I quit!