How to find the minimum of $abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$ when $ab+bc+cd+da+ac+bd=6$

Question

If $a,b,c,d\geq0$ are such that $ab+bc+cd+da+ac+bd=6$, then what is the minimum value of $$f(a,b,c,d)=abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$$ ?

My attempts: I think the minimum $5$ is obtained at $a=b=c=d=1$. I tried the Hölder inequality: $$f(a,b,c,d)\geq abcd + (1+\sqrt{a b c d})^2,$$ but the last term is $1<5$ for $d=0$.

Also, direct AM-GM on the two terms of $f$ did not work either. What to do?

Remark: This problem is from AoPS. There, it is also asked what happens for $ab+bc+cd+da+ac+bd=7$, which seems even harder.

Answer 1

We have the following identity:

$$\prod_{cyc} (a^2+1) = \left(\sum_{cyc}a-\sum_{cyc}abc\right)^2+\left(\sum_{sym}ab-abcd-1\right)^2$$

So

$$\prod_{cyc} (a^2+1) \geq \left(\sum_{sym}ab-abcd-1\right)^2$$

and this implies

$$1+abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} \geq ab+bc+cd+da+bd+ca$$

Now, we can see that the minimum is indeed $5$.

Answer 2

Preliminary remark. I just noticed that we can use the square root algorithm for polynomials to show in general that $$(a^2 + 1) (b^2 + 1) (c^2 + 1) (d^2 + 1) - (a b + b c + c d + d a + a c + b d - a b c d - 1)^2=(a + b + c +d- a b c - a b d - a c d - b c d)^2\geq0$$

which immediately implies that $f(a,b,c,d)+1\geq a b + b c + c d + d a + a c + b d$ for all $a,b,c,d\geq0$ with equality if and only if $$d=\frac{a+b+c-abc}{a b+a c+b c-1}.$$

But I noticed this too late, so here is some tedious manual work for you, which basically executes the above idea for the special cases $a b + b c + c d + d a + a c + b d=6$ and $7$.

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Tedious calculations:

Indeed the minimum is achieved at $5$. We need to prove $$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}\geq 5-abcd.$$ By direct AM-GM, we get $6=ab+bc+cd+da+ac+bd\geq 6\sqrt{abcd}$ so that $abcd\le1<5$. So we can square both sides of the inequality to prove and we now have to prove $$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\geq(5-abcd)^2.$$ Since $6=ab+bc+cd+da+ac+bd$, we get $d=\frac{6-a b-a c-b c}{a+b+c}$ and so we have to prove $$\left(a^2+1\right) \left(b^2+1\right) \left(c^2+1\right) \left(\frac{(-a b-a c-b c+6)^2}{(a+b+c)^2}+1\right)-\left(5-\frac{a b c (-a b-a c-b c+6)}{a+b+c}\right)^2\geq0$$

which is true because the left-hand side (amazingly!) equals $$\frac{\left(6 + a^2 - 5 a b + b^2 + a^2 b^2 - 5 a c - 5 b c + a^2 b c + a b^2 c + c^2 + a^2 c^2 + a b c^2 + b^2 c^2\right)^2}{(a+b+c)^2}.$$

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If we replace the condition by $7=ab+bc+cd+da+ac+bd$, a very similar approach works: Indeed I claim that the minimum of $f(a,b,c,d)$ is $6$ with that side constraint. Again by direct AM-GM, $a b c d \le \left(\frac{ab+bc+cd+da+ac+bd}6\right)^2=\frac{49}{36}<6$ so we can subtract $abcd$, square both sides, and need to prove $$(1 + a^2) (1 + b^2) (1 + c^2) (1 + d^2)-(6 - a b c d)^2\geq 0.$$

By the side constraint we have $d=\frac{7-a b-a c-b c}{a+b+c}$. Substituting this we need to prove $$\left(a^2+1\right) \left(b^2+1\right) \left(c^2+1\right) \left(\frac{(-a b-a c-b c+7)^2}{(a+b+c)^2}+1\right)-\left(6-\frac{a b c (-a b-a c-b c+7)}{a+b+c}\right)^2\geq 0.$$

But very nicely, the left-hand side can again be written as $$\frac{\left(7 + a^2 - 6 a b + b^2 + a^2 b^2 - 6 a c - 6 b c + a^2 b c + a b^2 c + c^2 + a^2 c^2 + a b c^2 + b^2 c^2\right)^2}{(a+b+c)^2}.$$

We have equality if and only if the side constraint is fulfilled and $$7 + a^2 - 6 a b + b^2 + a^2 b^2 - 6 a c - 6 b c + a^2 b c + a b^2 c + c^2 + a^2 c^2 + a b c^2 + b^2 c^2=0.$$

Note that this is a quadratic polynomial in each variable if the other variables are fixed. So for given $a$ and given $b$, we can solve for $c$. As an example, $a=1,b=1,c=\frac12$ leads to $d=2$ and indeed equality as $$f\left(1,1,\frac12,2\right)=6.$$

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Answer 3

For $a=b=c=d=1$ we obtain a value $5$.

We'll prove that it's a minimal value.

Indeed, let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$

$abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.

Thus, by AM-GM $v^4\geq t^4,$ $v^2=1$ and we need to prove that: $$abcd+\sqrt{a^2b^2c^2d^2+\sum\limits_{cyc}a^2b^2c^2+\frac{1}{4}\sum\limits_{sym}a^2b^2+\sum\limits_{cyc}a^2+1}\geq5$$ or $$t^4+\sqrt{t^8+(16w^6-12v^2t^4)v^2+(36v^4-32uw^3+2t^4)v^4+(16u^2-12v^2)v^6+v^8}\geq5v^4$$ or $$t^8-10v^4t^4+16v^2w^6+25v^8-32uv^4w^3+16u^2v^6\geq25v^8-10v^4t^4+t^8$$ or $$(uv^2-w^3)^2\geq0$$ and we are done!

Answer 4

Using the known identity $(u^2 + v^2)(x^2 + y^2) = (ux - vy)^2 + (uy + vx)^2$, we have $$(a^2 + 1)(b^2 + 1) = (ab - 1)^2 + (a + b)^2 \tag{1}$$ and $$(c^2+1)(d^2+1) = (cd - 1)^2 + (c + d)^2. \tag{2}$$

From (1) and (2), using Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)\\ =\,& [(ab - 1)^2 + (a + b)^2]\,[(cd - 1)^2 + (c + d)^2]\\ \ge\,& [(ab - 1)(cd - 1) - (a + b)(c + d)]^2\\ =\,& (abcd - ab - cd + 1 - ac - ad - bc - bd)^2 \end{align*} which results in $$\sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)} \ge ab + bc + cd + da + ac + bd - 1 - abcd.$$

Thus, we have $$abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)} \ge ab + bc + cd + da + ac + bd - 1 = 5.$$

Also, when $a = b = c = d = 1$, we have $abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)} = 5$.

Thus, the minimum of $abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}$ is $5$.