Minimize $P=\dfrac{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}{\sqrt{ab+bc+ca+6}},$ if $a+b+c+abc=4.$

Question

Problem. Let $a,b,c\ge 0: a+b+c+abc=4.$ Find minimal value of $P$ $$P=\frac{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}{\sqrt{ab+bc+ca+6}}.$$ Source: Vo Quoc Ba Can.

---

My attempt:

Set $$P(a,b,c)=\frac{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}{\sqrt{ab+bc+ca+6}}.$$

After some calculating works, I've tried prove $P(1,1,1)$ is the minimal value.

It means that we need to prove $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\ge\sqrt{2}\cdot\sqrt{ab+bc+ca+6}.$$ By squaring both side, it remains to prove $$a+b+c+3+2\sum_{cyc}\sqrt{(a+1)(b+1)}\ge 2(ab+bc+ca+6),$$or $$2\sum_{cyc}\sqrt{(a+1)(b+1)}\ge 5+abc+2(ab+bc+ca).$$ I was stucked here. Hope to see some ideas to continue the idea.

Also, all answer and comment are welcome. Thank for your attention.

Updated edit: Thank you @138 Aspen. I made a mistake in calculating.

Minimum should be $$P(0,2,2)=\frac{2\sqrt{30}+\sqrt{10}}{10}.$$

It means that we need to prove $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\ge\frac{2\sqrt{30}+\sqrt{10}}{10}\cdot\sqrt{ab+bc+ca+6},$$which seems ugly. Hope MV (mixing variables) technique works.

Answer 1

Two approaches.

Approach 1.

We may use Holder, similar to my answer here.

It suffices to prove that $$\sum_{\mathrm{cyc}}\sqrt{\frac{a + 1}{ab + bc + ca + 6}} \ge \frac{2\sqrt{30} + \sqrt{10}}{10}.$$

If $\max(a, b, c) = 4$, clearly the inequality is true. In the following, assume that $\max(a, b, c) < 4$.

By Holder, we have \begin{align*} &\Big(\sum_{\mathrm{cyc}}\sqrt{\frac{a + 1}{ab + bc + ca + 6}}\Big)^2 \sum_{\mathrm{cyc}} (a + 1)^2(ab + bc + ca + 6)(ab + \sqrt{3}\, bc + ca)^3\\[6pt] &\ge \left(\sum_{\mathrm{cyc}} (a + 1)(ab + \sqrt{3}\, bc + ca)\right)^3. \tag{1} \end{align*}

It suffices to prove that \begin{align*} &\left(\sum_{\mathrm{cyc}} (a + 1)(ab + \sqrt{3}\, bc + ca)\right)^3\\ \ge{}& \left(\frac{2\sqrt{30} + \sqrt{10}}{10}\right)^2 \sum_{\mathrm{cyc}} (a + 1)^2(ab + bc + ca + 6)(ab + \sqrt{3}\, bc + ca)^3. \tag{2} \end{align*} (2) is true which is verified by Mathematica - a Computer Algebra System.

$\phantom{2}$

Approach 2.

Fact 1. Let $p, q, r \ge 0$ with $p^2 \ge 3q$ and $r \ge \frac{4pq - p^3}{9}$ and $p^2 + 4pr + 2qr + r^2 + 2p - 2q + 8r - 4 = 0$. Then $$\frac{(3 + p)^2}{2pq - 2pr + q^2 + 4q - 6r + 6} \ge \frac{13 + 4\sqrt 3}{10}.$$ (It is verified by Mathematica (a Computer Algebra System).)

Let \begin{align*} p &:= (\sqrt{a+1} - 1) + (\sqrt{b + 1} - 1) + (\sqrt{c + 1} - 1), \\ q &:= \sum_{\mathrm{cyc}} (\sqrt{1 + a} - 1)(\sqrt{1 + b} - 1), \\ r &:= (\sqrt{1 + a} - 1)(\sqrt{1 + b} - 1)(\sqrt{1 + c} - 1). \end{align*} (Note: Letting $x = \sqrt{a+1} - 1$, $y = \sqrt{b + 1} - 1$ and $z = \sqrt{c + 1} - 1$, we have $p = x + y + z, q = xy + yz + zx, r = xyz$.)

Clearly, we have $p^2 \ge 3q$, and $r \ge \frac{4pq - p^3}{9}$ (degree three Schur). Also, we have $$p^2 + 4pr + 2qr + r^2 + 2p - 2q + 8r - 4 = a + b + c + abc - 4 = 0.$$ Moreover, we have $$\frac{(3 + p)^2}{2pq - 2pr + q^2 + 4q - 6r + 6} = \frac{(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})^2}{ab+bc+ca+6}.$$ By Fact 1, we have $$ \frac{(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})^2}{ab+bc+ca+6} \ge \frac{13 + 4\sqrt 3}{10}.$$

Answer 2

It means that we need to prove $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\ge\frac{2\sqrt{30}+\sqrt{10}}{10}\cdot\sqrt{ab+bc+ca+6},$$

Put $M=\left(\frac{2\sqrt{30}+\sqrt{10}}{10}\right)^2$, $x=\sqrt{a+1}\ge 1$, $y=\sqrt{b+1}\ge 1$, and $z=\sqrt{c+1}\ge 1$.

Thus we need to show $(x+y+z)^2\ge M(ab+bc+ca+6)$.

We have $$x^2+y^2+z^2=a+b+c+3,$$ $$x^2y^2z^2=(a+1)(b+1)(c+1)=$$ $$abc+ab+bc+ca+a+b+c+1=ab+bc+ca+5,$$ $$x^2y^2+y^2z^2+z^2x^2=$$ $$(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)=$$ $$ab+bc+ca+2(a+b+c)+3=$$ $$ab+bc+ca+2(x^2+y^2+z^2)-3.$$

Thus we have to show that

$$(x+y+z)^2\ge M(x^2y^2+y^2z^2+z^2x^2-2(x^2+y^2+z^2)+9)$$

provided $x,y,z\ge 1$ and $$x^2y^2z^2-5=x^2y^2+y^2z^2+z^2x^2-2(x^2+y^2+z^2)+3.$$

Maybe this can be achieved using Lagrange's multipliers.

Answer 3

We have for $x\in[0,2]$ :

$$\sqrt{x+1}\geq g(x)=\frac{\left(1-\sqrt{3}\right)}{0-2}x+1+\frac{1}{8}\frac{x\left(2-x\right)}{x+2}+\frac{1}{3}\frac{x^{2}\left(2-x\right)}{\left(x+2\right)^{4}}$$

Then :

$$a(x,y,z)=(g(x)+g(y)+g(z))^{2}-\left(\frac{2\sqrt{30}+\sqrt{10}}{10}\right)^2(xy+yz+zx+6)$$

Then it seems that all the coefficient are positive in :

$$a\left(\frac{2x}{x+1},\frac{2y}{y+1},\frac{2(x+y+2)}{5xy+x+y+1}\right)$$

Ps: it's painful need a software better than Geogebra.

If you want to check it's all positive .(false)

Another way :

We have for $x\in[0,2]$ :

$$\sqrt{x+1}\geq g(x)=\frac{\left(1-\sqrt{3}\right)}{0-2}x+1+\frac{1}{8}\frac{x\left(2-x\right)}{x+2}+\frac{1}{\left(16\left(7+4\sqrt{3}\right)\right)}x\left(2-x\right)$$

We have for $x\in[3,4]$ :

$$f\left(x\right)=\sqrt{x+6}\leq h\left(x\right)=\frac{\left(x+6\right)}{\left(f\left(4\right)-f\left(3\right)\right)\left(x-4\right)+f\left(4\right)}$$

Then it seems true that :

$$a(x,y,z)=(g(x)+g(y)+g(z))-\left(\frac{2\sqrt{30}+\sqrt{10}}{10}\right)h\left(xy+yz+zx\right),a\left(\frac{2x}{x+1},\frac{2y}{y+1},\frac{2(x+y+2)}{5xy+x+y+1}\right)\geq 0$$

Can someone check (I check it's positive for $x\geq 1,y\geq 1$ as we have some constraint on the bound above):

$$a\left(\frac{2x}{x+1},\frac{2y}{y+1},\frac{2(x+y+2)}{5xy+x+y+1}\right)>0$$

Using Am-Gm it's not hard to show it's positive for $x\geq 1,y\geq 1$ it's a bit lenghty .

Using Jensen's inequality :

define :

$$g(x)=\frac{\left(1-\sqrt{3}\right)}{0-2}x+1+\frac{1}{8}\frac{x\left(2-x\right)}{x+2}+\frac{1}{\left(16\left(7+4\sqrt{3}\right)\right)}x\left(2-x\right),t\left(x\right)=\left(x+2+\frac{1}{x+d}\right)g\left(x\right),h\left(x\right)=\left(x+2+\frac{1}{x+d}\right)\left(g\left(x\right)-\frac{t''\left(0\right)x^{2}}{2\left(x+2+\frac{1}{x+d}\right)}\right)$$

Such that : $$h''(0)=h''(2)=0,d\simeq 6$$

Then we have $x,y\in[1,\infty)$:

$$d\left(x,y\right)>b\left(x,y\right)$$

Where :

$$l\left(x,y,z\right)=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}-\frac{\left(2\sqrt{30}+\sqrt{10}\right)}{10}\left(\sqrt{6+xy+yz+zx}\right),d\left(x,y\right)=l\left(\frac{2x}{x+1},\frac{2y}{y+1},\frac{2(x+y+2)}{5xy+x+y+1}\right)$$

And :

$$a\left(x,y,z\right)=\left(\frac{1}{m\left(x\right)}+\frac{1}{m\left(y\right)}+\frac{1}{m\left(z\right)}\right)\left(h\left(\frac{\left(\frac{x}{m\left(x\right)}+\frac{y}{m\left(y\right)}+\frac{z}{m\left(z\right)}\right)}{\frac{1}{m\left(x\right)}+\frac{1}{m\left(y\right)}+\frac{1}{m\left(z\right)}}\right)\right)-\frac{\left(2\sqrt{30}+\sqrt{10}\right)}{10}\left(xy+yz+zx+6\right)\cdot\frac{1}{\left(\sqrt{10}-3\right)\left(xy+yz+zx-4\right)+\sqrt{10}}+\frac{j\left(x\right)}{m\left(x\right)}+\frac{j\left(y\right)}{m\left(y\right)}+\frac{j\left(z\right)}{m\left(z\right)},m\left(x\right)=\left(x+2+\frac{1}{x+6}\right),b\left(x,y\right)=a\left(\frac{2x}{x+1},\frac{2y}{y+1},\frac{2(x+y+2)}{5xy+x+y+1}\right),j\left(x\right)=\frac{t''\left(0\right)x^{2}}{2}$$

Answer 4

If you trust in computer based proofs, this can be done with Maple >=2018 as follows. First, we numerically solve the optimization problem

Optimization:-Minimize((sqrt(a+1)+sqrt(1+b)+sqrt(1+c))/sqrt(a*b+a*c+b*c+6), {a >= 0, b >= 0, c >= 0, a*b*c+a+b+c = 4});

$$[ 1.41421356237309515,[a= 1.0,b= 1.0,c= 1.0]] $$

The above result suggests that the minimum $\sqrt 2$ is attained at $(a=1,b=1,c=1)$.

To check it, we eliminate $a$ from the target function by

solve(a*b*c+a+b+c = 4, a);

-(-4+b+c)/(b*c+1)

f := eval((sqrt(a+1)+sqrt(1+b)+sqrt(1+c))/sqrt(a*b+a*c+b*c+6), a = -(-4+b+c)/(b*c+1));

f := (sqrt(-(-4+b+c)/(b*c+1)+1)+sqrt(1+b)+sqrt(1+c))/sqrt(-(-4+b+c)*b/(b*c+1)-c*(-4+b+c)/(b*c+1)+b*c+6)

Third, we use the second derivative test with Maple by the command

Student:-MultivariateCalculus:-SecondDerivativeTest(f, [b, c] = [1, 1]);

LocalMin = [[1, 1]], LocalMax = [], Saddle = []

Of course, all that may be mimicking by hand.

Addition. As I was informed by @V.S.e.H., this is a local minimum only. A global minimum can be found with

DirectSearch:-GlobalOptima((sqrt(a+1)+sqrt(1+b)+sqrt(1+c))/sqrt(a*b+a*c+b*c+6), {a >= 0, b >= 0, c >= 0, a*b*c+a+b+c = 4});

[1.41167288061134, [a = 1.99999839031817, b = 3.18835950887659*10^(-13), c = 2.00000161802040], 2167]

The result suggests that three global minimums are attained at the corners of the triangle (the feasible set). Now we investigate the behavior of the target function on the side by usual tools of calculus with Maple.

eval((sqrt(a+1)+sqrt(1+b)+sqrt(1+c))/sqrt(a*b+a*c+b*c+6), {b = 0, c = 4-a});

(sqrt(a+1)+1+sqrt(5-a))/sqrt(6+a*(4-a))

minimize((sqrt(a+1)+1+sqrt(5-a))/sqrt(6+a*(4-a)), a = 0 .. 4, location);

(1/10)*(2*sqrt(3)+1)*sqrt(10), {[{a = 2}, (1/10)*(2*sqrt(3)+1)*sqrt(10)]}

There is no need to consider other sides because of symmetry. I repeat all that may be mimicked by hand.

Answer 5

Remarks: I found a proof better than my two old approaches. We use the so-called isolated fudging. For my answers using isolated fudging, see e.g. P1, P2, P3.

We need to prove that $$\sqrt{a + 1} + \sqrt{b + 1} + \sqrt{c + 1} \ge \frac{2\sqrt{30} + \sqrt{10}}{10} \cdot \sqrt{ab + bc + ca + 6}. \tag{1}$$

Let $C := \big(\frac{2\sqrt{30} + \sqrt{10}}{10}\big)^2$. Squaring both sides of (1), it suffices to prove that $$\sum_{\mathrm{cyc}} 2\sqrt{a + 1}\sqrt{b + 1} \ge C(ab + bc + ca + 6) - a - b - c - 3. \tag{2}$$

Let \begin{align*} f(a, b, c) &:= \frac{\sqrt{3} + 2}{5}abc + \frac{129 - 28\sqrt{3}}{90}ab + \frac{16\sqrt{3} - 3}{45}(a + b)c\\[6pt] &\qquad + \frac{14\sqrt{3} + 3}{45}(a + b) + \frac{3 - \sqrt 3}{45}c. \end{align*} We have \begin{align*} &C(ab + bc + ca + 6) - a - b - c - 3 - \sum_{\mathrm{cyc}} f(a, b, c)\\ ={}& - \frac35(2 +\sqrt 3)(a + b + c + abc - 4)\\[6pt] ={}& 0. \tag{3} \end{align*}

Using (3), (2) is written as $$\sum_{\mathrm{cyc}} \left(2\sqrt{a + 1}\sqrt{b + 1} - f(a, b, c)\right) \ge 0. \tag{4}$$

Fact 1. Let $x, y, z \ge 0$ with $x + y + z + xyz = 4$. Then $$2\sqrt{x + 1}\sqrt{y + 1} - f(x, y, z) \ge 0.$$ (The proof is given at the end.)

By Fact 1, letting $x = a, y = b, z = c$, we have $2\sqrt{a + 1}\sqrt{b + 1} - f(a, b, c) \ge 0$; letting $x = b, y = c, z = a$, we have $2\sqrt{b + 1}\sqrt{c + 1} - f(b, c, a)\ge 0$; letting $x = c, y = a, z = b$, we have $2\sqrt{c+1}\sqrt{a+1} - f(c, a, b) \ge 0$. Thus, (4) is true.

We are done.

$\phantom{2}$

---

Proof of Fact 1.

We need to prove that \begin{align*} 2\sqrt{xy + x + y + 1} &\ge \frac{\sqrt{3} + 2}{5}xyz + \frac{129 - 28\sqrt{3}}{90}xy + \frac{16\sqrt{3} - 3}{45}(x + y)z\\[6pt] &\qquad + \frac{14\sqrt{3} + 3}{45}(x + y) + \frac{3 - \sqrt 3}{45}z. \tag{A1} \end{align*}

Let $p = x + y, q = xy$. Then $0 \le p \le 4$, $q \ge 0$ and $p^2\ge 4q$. Using $z = \frac{4 - x - y}{1 + xy} = \frac{4 - p}{1 + q}$, (A1) is equivalently written as \begin{align*} 180(1 + q)\sqrt{p + q + 1} &\ge ( 129-28\sqrt {3} ) {q}^{2} + ( 10\sqrt {3}\,p+44 \sqrt {3}-30p+273 ) q\\ &\quad -32\,\sqrt {3}{p}^{2}+158\,\sqrt {3}p+6\, {p}^{2}-8\,\sqrt {3}-24\,p+24.\tag{A2} \end{align*} Let $F(q) := \mathrm{LHS}_{(A2)} - \mathrm{RHS}_{(A2)}$. We have, for all $0 \le q \le p^2/4$, \begin{align*} F''(q) &= \frac{45(4p + 3q + 3)}{(p + q + 1)^{3/2}} - 258 + 56\sqrt{3} \\ &\le \frac{45(4p + 3\cdot p^2/4 + 3)}{(p + 1)^{3/2}} - 258 + 56\sqrt{3}\\ &< 0. \end{align*} Thus, $F(q)$ is concave on $[0, p^2/4]$. Also, we have $F(0) \ge 0$ and $F(p^2/4) \ge 0$ (easy), for all $0\le p \le 4$. Thus, $F(q) \ge 0$ on $[0, p^2/4]$, given any $0 \le p\le 4$.

We are done.

Answer 6

Let us rewrite. We are searching the minimal value of the expression $$ P(a,b,c)= \frac {\sqrt{a+1} + \sqrt{b+1} + \sqrt{c+1}} {\sqrt{(a+1)(b+1)(c+1)+1}} $$ constrained to the relation $a+b+c+abc=4$. It turns out it is $\sqrt K$, $$ P(a,b,c)\ge \sqrt K\ ,\qquad K=\frac1{10}(1+2\sqrt 3)^2=\frac 1{10}(13+2\sqrt 7)\approx 1.9928203230275\dots\ \ . $$ Let us show this.

---

It is convenient to substitute $x,y,z\ge 1$ for $\sqrt{a+1}$, $\sqrt{b+1}$, $\sqrt{c+1}$. And let us use the symmetric elementary functions $p,q,r$ for the variables $x,y,z$, so $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. The relation satisfied by $x,y,z$ is: $$ x^2y^2z^2 -x^2y^2 -y^2z^2-z^2x^2 +2x^2+2y^2+2z^2-8=0\ . $$ In terms of $p,q,r$ it is: $$ r^2 -(q^2-2pr)+2(p^2-2q)-8=0\ . $$ So we solve a related problem first, we minimize the following function $F$ on the domain $p\ge 3$, $q\ge 3$, $r\ge 1$, $p^2\ge 3p$ (which is $\sum(x-y)^2\ge 0$ with equality iff $x=y=z$), and $p(p^2-3q) + 6r\ge pq-3r$ (which is $x^3+y^3+z^3+3xyz\ge x^2(y+z)+y^2(z+x)+z^3(x+y)$, Schur). Note that the Schur inequality becomes an equality if and only if two involved variables are equal, this is also the case we obtain equality in $P\ge \sqrt K$. We restate this as $P^2\ge K$, and further as $(x+y+z)^2\ge K(x^2y^2z^2+1)$. So we consider the minimization of $F$, $$ F(p,q,r):=p^2-K(r^2+1)\ , \qquad\text{ constrained to } \underbrace{r^2-(q^2-2pr)+2(p^2-2q)-8}_{h(p,q,r)}=0\ . $$

---

Lagrange multiplicators. In case of a local extremal value in the domain (two-dimensional) $h(p,q,r)=0$, with boundaries given by the above boundary conditions, the coresponding extremal point satisfies $0=G'_p=G'_q=G'_r=G'_m$ for $$ G(p,q,r;m):= F(p,q,r)-mh(p,q,r)\ . $$ The corresponding system is: $$ \left\{ \begin{aligned} 0&=2p-m(2r+4p)\ ,\\ 0&=-m(-2q-4)\ ,\\ 0&=-2Kr-m(2r+2p)\ ,\\ 0&=h(p,q,r)\ . \end{aligned} \right. $$ The condition $0=G'_q$ makes everything clear. ($F$ does not depend on $q$.) The factor $2q+4$ does not vanish in the given region. Also, $m=0$ leads to $p=r=0$, again outside of the sight. So any extremal value for $F$ constrained to $h=0$ is a boundary point.

Each of the boundary conditions $p=3$ and/or $q=3$ and/or $r=1$ and/or $p^2=3q$ leads to $x=y=z$ thus to $a=b=c$, and the common value is $1$ from the given $(a,b,c)$-condition. In this case $x=y=z=\sqrt 2$, $p=3\sqrt 2$, $r=2\sqrt 2$, and $$ \begin{aligned} F(3\sqrt 2,*,2\sqrt 2) &=(3\sqrt 2)^2-K((2\sqrt2)^2+1)=18-9K=9(2-K) \\ &= \frac 9{10} (7-4\sqrt 3) =\frac 9{10} \cdot \frac {49-48}{7+4\sqrt 3} >0\ .e \end{aligned} $$

---

It remains to check the remained boundary condition, the Schur condition on $p,q,r$, taken with equality. This happens iff two values among $x,y,z$ (equivalently among $a,b,c$) are equal. Here, we can go back to $P$, assume $a=c$, write $b$ in terms of $a$ as $b=b(a)=2(2-a)/(a^2+1)$, so $a\in[0,2]$, and minimize the ugly function in one variable

$$a\to g(a)=\Bigg( \ 2\sqrt{a+1} +\sqrt{b(a)+1}\ \Bigg)^2-K\Bigg(\ 1 + (a+1)\sqrt{b(a)+1} \ \Bigg)\ , \qquad g:[0,2]\to\Bbb R\ . $$ In a plot:

I hope, everybody who considered the problem will find from here her or his way to proceed.

---

Humanly, the problem starts now, i have no chance to finish in time as a human. The reason is the following one. We have a more or less complicated function. Its study must somehow involve analytic tools like taking derivatives, consider critical points, compute the value in these points. Then also compute the values at the ends of the interval $[0,2]$, and finally decide which candidate wins the minimum prize.

Well, this consideration must then find the critical point $a=1$, which is easy to check in any system of equations, but also that local max between $1$ and $ 2$. And this point is no longer "easy". So let us work algebraically with the aid of a computer. We are searching for globally minimal points $(a,b,c)$ for $P$ with $a=c$. Let us pass to the $(x,y,z)$ coordinates via $a=x^2-1$, $b=y^2-1$, the condition $a=c$ becomes $x=z$, so we want to minimize $$ \begin{aligned} F(x,y ) &= (2x+y)^2-K(x^4y^2+1)\ ,\\ &\qquad\text{ constrained to } \\ 0 &= \underbrace{2(x^2-1)+(y^2-1)+(x^2-1)^2(y^2-1)-4}_{h(x,y)}\ . \end{aligned} $$ We have two new functions $F,h$. Lagrange multiplicators for them. We obtain an ugly algebraic system of high degree. Let us ask the machine for solutions with $x,y\ge 1$.

L.<w> = QuadraticField(3)
K = (1 + 2*w)^2/10
R.<x, y, m> = PolynomialRing(L)

F = (2*x + y)^2 - K*(x^4*y^2 + 1)
h = 2*(x^2 - 1) + (y^2 - 1) + (x^2 - 1)^2*(y^2 - 1) - 4

J = R.ideal([diff(F - m*h, v) for v in (x, y, m)])
variety = J.variety(ring=AA)
for dic in variety:
    x0, y0, m0 = dic[x], dic[y], dic[m]
    if not(x0 >= 1 and y0 >= 1):
        continue
    a0, b0 = x0^2 - 1, y0^2 -1
    print(f"Critical point with components:")
    print(f"(x, y) = ({x0}, {y0})")
    print(f"(a, b) = ({a0}, {b0})")
    print(f"    F-value = {F.subs(x=x0, y=y0)}")

And we obtain:

Critical point with components:
(x, y) = (1.414213562373095?, 1.414213562373095?)
(a, b) = (1, 1)
    F-value = 0.06461709275204174?
Critical point with components:
(x, y) = (1.610339275596270?, 1.109030006353267?)
(a, b) = (1.593192582527918?, 0.2299475549919257?)
    F-value = 0.2710105546461977?

This values are now compared with the values at the extremities of the interval, and the value in $(a,b)=(2,0)$, corresponding to $(x,y)=(\sqrt 3,1)$ wins, the function $F$ is zero, the global minimum.