# Proving $abcd+3\geq a+b+c+d$

If $$a,b,c,d$$ are non negatives and $$a^2+b^2+c^2+d^2=3$$ prove that $$abcd+3\ge a+b+c+d$$

The inequality is not as simple as it looks.The interesting part is that the equality occurs when $$a=0,b=c=d=1$$ upto permutation .(I don't know if there are further equality cases.)

I tried rewriting the inequality as $$a^2+a(bcd-1)+b^2+c^2+d^2-b-c-d\ge 0$$ As its a quadratic in $$a$$ it suffices to show the discriminant$$\Delta_a={(bcd-1)}^2-4(b^2+c^2+d^2-b-c-d)\le 0$$ which unfortunately is wrong (when $$a=\sqrt{3},b=c=d=0$$)

P.S ;I am not aware of using Lagranges multipliers

Its from here

• I haven't figured out how to use that but in post #6 arqady (I guess a.k.a. @MichaelRozenberg at MSE) seems to give a hint... – VIVID Jan 18 at 14:30
• @VIVID nice find! thanks for your interest – Albus Dumbledore Jan 18 at 14:34
• As the problem is symmetric with respect to all variables, I tried to take $a=b=c=d=\frac{\sqrt 3}{2}$ and got $$a b c d-(a+b+c+d)\ge \frac{9}{16}-2 \sqrt{3}\approx -2.9$$ which is the exact minimum. I don't write this as an answer because without analysis I can't prove that the minimum happens when the variables are equal. – Raffaele Jan 18 at 15:39
• @Raffaele interesting ,that means this inequality is infact weaker...... – Albus Dumbledore Jan 18 at 15:49
• -- FYI -- The statement holds more generally for three, or more than four real variables without positivity assumption: $$\text{If \,3\leqslant n\, and}\;\sum^n_{d=1}a_d^2=n-1, \text{ then }\:\sum^n_{d=1}a_d\:\leqslant\:\prod^n_{d=1}a_d+n-1,$$ cf page 57 in the GAZETA MATEMATICA . Notice that $\,a_d\,$ encodes Albus Dumbledore ... – Hanno Jan 21 at 16:11

Due to symmetry, assume wlog that $$a\geqslant b\geqslant c\geqslant d\geqslant 0$$. Since I will use this fact later, I am proving it before the cases:

Lemma: We have that $$2\geqslant a+d$$.

In fact, this follows from Cauchy-Schwarz: $$2= 2\sqrt{\frac{a^2+b^2+c^2+d^2}{3}}\geqslant 2\sqrt{\frac{a^2+3d^2}{3}}=\sqrt{1+\frac13}\cdot \sqrt{a^2+3d^2}\geqslant a+d$$

We will now consider some cases depending on the position of the number $$1$$:

• Case 1: $$1\geqslant a\geqslant b\geqslant c\geqslant d\geqslant 0$$. Observe that one may rewrite the inequality as $$3+abcd-(a+b+c+d)=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd)$$ Which is clearly positive.

• Case 2: $$a\geqslant 1\geqslant b\geqslant c\geqslant d\geqslant 0$$. If $$ab\geqslant 1$$, we might proceed as in the next case. So we will work with $$ab<1$$. Thus \begin{align*}3+abcd-(a+b+c+d)&=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd)\\&\geqslant \underbrace{(1-a)}_{<0}(1-b)+(1-c)(1-d)\\&\geqslant (1-a)(1-c)+(1-c)(1-d)\\&=(1-c)(2-(a+d))\geqslant 0 \end{align*} Where the last inequality follows from the lemma.

• Case 3: $$a\geqslant b\geqslant 1\geqslant c\geqslant d\geqslant 0$$. This implies that \begin{align*}6+2abcd-2(a+b+c+d)&=a^2+b^2+c^2+d^2+3+2abcd-2(a+b+c+d)\\ &=(a-1)^2+(b-1)^2+(c+d-1)^2+2cd(ab-1)\geqslant 0\end{align*} Or, equivalently $$3+abcd\geqslant a+b+c+d$$.

• Case 4: $$a\geqslant b\geqslant c\geqslant 1\geqslant d\geqslant 0$$. As @dezdichado noticed, this case is straightforward, since it forces directly $$a=b=c=1, d=0$$ due to the constraint $$a^2+b^2+c^2+d^2=3$$.

• Case 5: $$a\geqslant b\geqslant c\geqslant d\geqslant 1$$. Clearly impossible, since this would yield $$a^2+b^2+c^2+d^2\geqslant 4>3$$.

Done!

• your case $3$ would have forced $a = b= c = 1$ and $d = 0$ anyway due to the constraint. +1 – dezdichado Jan 18 at 16:06
• Ingenious(+1) I doubt a better answer will be there – Albus Dumbledore Jan 18 at 16:12
• I don't think so? E.g. $a=\sqrt{3}, b=c=d=0.$ Then L.H.S. $=0$ but R.H.S. =$\sqrt{3}+1$. – See Hai Jan 18 at 16:20
• @AlbusDumbledore I also miss Michael Rozenberg; he's able to solve (almost) every inequality, and some of them with crazy computations (such as BW or uvw ones)!! It's sad that he does not answer questions anymore on this site (does anybody now why?). Anyway, he still comes and looks around, since he's been active at least yesterday. Hope he comes back – Dr. Mathva Jan 18 at 16:48
• @Dr.Mathva :) He is still answering questions in AOPS, I am a fan of him too! – Albus Dumbledore Jan 18 at 16:52

Since $$a^2+b^2+c^2+d^2 = 3$$ defines a compact set in $$\mathbb{R}^4$$, $$f(a,b,c,d)=abcd+3-a-b-c-d$$ will have a global minimum and maximum over that set, that will occur (can be easily justified) in critical points of the Lagrangian $$L(a,b,c,d,\lambda) = abcd+3-a-b-c-d -\lambda(a^2+b^2+c^2+d^2-3)$$

So, just compute the critical points, compute the value of $$f$$ over each of these points and get the global minimum. If the global minimum is $$\ge 0$$, you are done.

When computing the critical points, you will get some complex solutions, some solutions with negative components but, in the end, the relevant solutions are the ones you already mentioned and also $$a=b=c=d=\frac{\sqrt{3}}{2}$$. The minimum value of $$f$$ (even admitting negative values for $$a,b,c,d$$) is zero (and the maximum is $$\frac{57}{16}+2\sqrt{3}$$).

Edit: Missed one critical point, but the result stands.

• sorry but what is a compact set and langrangian thing i am not at all aware – Albus Dumbledore Jan 18 at 14:11
• @AlbusDumbledore A compact set in $\mathbb{R}^4$ is a bounded and closed set. THis is relevant because a continuous function ($f$) always attains a global maximum and minimum values over a compact set. – PierreCarre Jan 18 at 14:14
• IHMO, the bulk o f Lagrange multiplier approach is always working on the derivatives so as to find the global extrema, which your answer didn't shed any light on. Your answer might just be a comment as VIVID posted. – Quang Hoang Jan 18 at 14:24
• My point is, yes, one doesn't need to provide all the calculations, but your answer is simply try Lagrangian Multiplier. It doesn't provide any further information specific to this question other than the generic method. E.g. Is it a straight forward process as solving a quadratic polynomial or what should OP expect from using this method to this question? – Quang Hoang Jan 18 at 14:32
• @QuangHoang the solution is fine. Smooth function on a compact set attains its maximum and minimum in that set and they can only be achieved at a critical point. Then, you use Lagrange to find all the critical points and then find the maximum and minimum and by just looking at their values. Your concern would be valid if there happen to be an infinite number of critical points, so inspecting them individually would be hard, but it's not the case here assuming his computation is right. – dezdichado Jan 18 at 15:44

This is a remark on Dr. Mathva's answer, which may seem very mysterious at first but becomes more transparent when you consider a simpler problem:

If $$a, b \geq 0$$ with $$a^2 + b^2 = 1$$, prove that $$ab+1 \geq a+b \,.$$

(It is the problem from the question if you impose $$c = d = 1$$.)

Here it is clear that you can factorize it as $$(a-1)(b-1) \geq 0$$, so you are naturally led to distinguish cases according to whether $$a, b \geq 1$$ or $$\leq 1$$. If both are $$\geq 1$$ or both are $$\leq 1$$, we are done. So suppose wlog $$a \geq 1 \geq b$$. Then plug in the constraint to get $$ab + a^2+b^2 \geq a+b \,.$$ Now play a bit to recover every term in the RHS from the LHS. Because $$a \geq 1$$, we can just remove factors $$a$$ from the LHS. That is, $$ab \geq b$$ and $$a^2 \geq a$$, and we are done.

By considering this simpler problem, you can more easily get the idea of assuming wlog $$a \geq b \geq c \geq d$$, and of distinguishing cases according to the position of $$1$$.

• Nice (+1) gives some intuition on Dr Mathvas answer – Albus Dumbledore Jan 20 at 10:17

See, in the link below, our extension to n real variables (GMA Problem 494, page 57, with the our solution): https://ssmr.ro/gazeta/gma/2020/gma3-4-2020-continut.pdf