Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$

Question

Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$

What I have tried so far is using CBS:

$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$

$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$

$(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$

Then, we have:

$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$.

Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.

Then, I used the following substitution:

$c + d = t$ and $a + b = 4 - t$.

We assume wlog that $a \leq b \leq c \leq d$.

Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.

Then, what we have to prove is:

$9t^2(4 - t)^2 \geq 256$.

We can rewrite this as:

$(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or $(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,

at which point I got stuck.

Yes, I agree, so either the second inequality I used by considering the order of a,b,c,d, or the CBS in the first place are not strong enough to solve the problem, but I had no other idea on how to solve it. — Sandel, Sep 02 '18 at 14:03

farruhota · Answer 1 · 2018-09-02T19:10:58.847

Minimize $f(a,b,c,d)=(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3)$ subject to $a+b+c+d=4$.

The Lagrange function: $L(a,b,c,d,t)=f(a,b,c,d)+t(4-a-b-c-d)$.

FOC: $$\begin{cases}L_a=2a(b^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\ L_b=2b(a^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\ L_c=2c(a^2 + 3)(b^2 + 3)(d^2 + 3)-t=0\\ L_d=2d(a^2 + 3)(b^2 + 3)(c^2 + 3)-t=0\\ L_t=4-a-b-c-d=0\\ \end{cases}$$ Consider the difference: $$L_a-L_b=(c^2+3)(d^2+3)(2ab^2+6a-2ba^2-6b)=0 \iff \\ (2ab-6)(b-a)=0 \Rightarrow 1) \ a=b; \ \ 2) \ ab=3.$$ Similarly, other differences are found: $$1)\ a=c; \ \ 2) \ ac=3\\ 1)\ a=d; \ \ 2) \ ad=3\\ 1)\ b=c; \ \ 2) \ bc=3\\ 1)\ b=d; \ \ 2) \ bd=3\\ 1)\ c=d; \ \ 2) \ cd=3\\$$ Cases: $$\begin{align} \ &1) \ a=b=c\ne d \Rightarrow ad=bd=cd=3 \Rightarrow 4-\frac 3d-\frac 3d-\frac 3d-d=0 \Rightarrow \emptyset;\\ &2) \ a=b\ne c=d \Rightarrow ac=ad=bc=bd=3 \Rightarrow 4-\frac 3d-\frac3d-d-d=0 \Rightarrow \emptyset \\ &3) \ a=b\ne c\ne d\ne a \Rightarrow ac=bc=cd=ad=bd=3 \Rightarrow 4-\frac 3d-\frac3d-\frac3d-d=0 \Rightarrow \emptyset; \\ &4) \ a=b=c=d \Rightarrow a+b+c+d=4 \Rightarrow a=b=c=d=1.\end{align}$$ The only solution is: $(a,b,c,d)=(1,1,1,1)$.

Now we will check bordered Hessian (let $g(x)=a+b+c+d$): $$\bar{H}=\begin{vmatrix} 0&g_a&g_b&g_c&g_d\\ g_a&L_{aa}&L_{ab}&L_{ac}&L_{ad}\\ g_b&L_{ba}&L_{bb}&L_{bc}&L_{bd}\\ g_c&L_{ca}&L_{cb}&L_{cc}&L_{cd}\\ g_d&L_{da}&L_{db}&L_{dc}&L_{dd}\\ \end{vmatrix}=\begin{vmatrix} 0&1&1&1&1\\ 1&128&64&64&64\\ 1&64&128&64&64\\ 1&64&64&128&64\\ 1&64&64&64&128\\ \end{vmatrix} \Rightarrow \\ \bar{H}_1=-1<0; \ \bar{H}_2=-128<0; \ \bar{H}_3=-12288<0; \ \bar{H}_4=-1048576<0,$$ which implies $f(1,1,1,1)=256$ is minimum.

@Michael Rozenberg, I updated to show the full solution. Hope did not miss anything. Regards. — farruhota, Sep 02 '18 at 17:21
The inequality $a+b+c+d\geq4\sqrt[4]{abcd}$ is wrong because our variables can be negative. — Michael Rozenberg, Sep 02 '18 at 17:34
@Michael Rozenberg, yes, you are right, keep forgetting it, fixed. thank you — farruhota, Sep 02 '18 at 17:42
OK. Now you need to explain, why this critical point is a minimal point and to tell us about some property of the continuous function on the compact. For me all this is very ugly, and does not help for hard inequalities. But thank you for your work! — Michael Rozenberg, Sep 02 '18 at 18:33
@Michael Rozenberg, completed by checking the stationary point, which is minimum. Thank you for collaboration. — farruhota, Sep 02 '18 at 19:12

score 4 · Answer 2 · edited Jul 23 '19 at 19:31

We have \begin{align} (a^2+3)(b^2+3) & =(a^2+1+2)(b^2+1+2) \\ & =(a^2+1)(b^2+1)+2(a^2+1+b^2+1)+4 \\ & = (a^2+1)(b^2+1)+2(a^2+b^2+2)+4 \\ & =(a^2+1)(b^2+1)+2(a^2+b^2)+8 \end{align} By Cauchy Schwartz we obtain $$(a^2+1)(b^2+1)\geq(a+b)^2\tag{$\star$}$$ $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(a+b)^2\tag{$\star \star$}$$ Puting together ($\star$) and ($\star$) we get $$(a^2+3)(b^2+3)=(a^2+1)(b^2+1)+2(a^2+b^2)+8\geq 2(a+b)^2+8=2[(a+b)^2+4]\tag{1}$$ In the same way we get $$(c^2+3)(d^2+3)=(c^2+1)(d^2+1)+2(c^2+d^2)+8\geq 2(c+d)^2+8=2[(c+d)^2+4]\tag{2}$$ Multiplying (1) and (2) we get the conclusion \begin{align} (a^2+3)(b^2+3)(c^2+3)(d^2+3) & \ge 4[(a+b)^2+4][(c+d)^2+4] \\ & \ge 4[2(a+b)+2(c+d)]^2 \tag{Cauchy Schwartz}\\ & =16(a+b+c+d)^2 =256 \end{align}

Michael Rozenberg · Answer 3 · 2018-09-02T15:26:26.617

3

We need to prove that $$\sum_{cyc}\ln(a^2+3)\geq4\ln4$$ or $$\sum_{cyc}\left(\ln(a^2+3)-\ln4-\frac{1}{2}(a-1)\right)\geq0.$$ Now, let $f(x)=\ln(x^2+3)-\ln4-\frac{1}{2}(x-1).$

Thus, $$f'(x)=\frac{(x-1)(3-x)}{2(x^2+3)},$$ which says that $f$ decreases on $[3,+\infty)$ and since $f(3)>0$, there is unique $x_0>3$, for which $f(x_0)=0$.

And indeed, $x_0=4.586...$ and since $f(1)=0$, our inequality is proven for $\max\{a,b,c,d\}\leq x_0$.

Let $a>x_0$.

Thus, $$\prod_{cyc}(a^2+3)\geq(x_0^2+3)\cdot3^3>256$$ and we are done!

edited Sep 02 '18 at 15:26

answered Sep 02 '18 at 15:13

Michael Rozenberg

203,855

I'm sorry, what means $cyc$ under the sum? – Severin Schraven Sep 02 '18 at 15:15
@Severin Schraven $\sum\limits_{cyc}\ln(a^2+3)=\ln(a^2+3)+\ln(b^2+3)+\ln(c^2+3)+\ln(d^2+3).$ – Michael Rozenberg Sep 02 '18 at 15:16
1

Is this a standard notation? I have not seen this before. – Severin Schraven Sep 02 '18 at 15:19
@Severin Schraven Yes, it's a standard notation. See here: https://artofproblemsolving.com/wiki/index.php?title=Cyclic_sum – Michael Rozenberg Sep 02 '18 at 15:20
trying to understand, if $x_0=4.586...$, can $a>x_0$ (what about the constraint $a+b+c+d=4$)? if $\max{a,b,c,d}\le x_0$, can $a>x_0$ be true? can the inequality be proved without derivatives and by CS, Schur, Muirhead, etc? – farruhota Sep 02 '18 at 16:13
@farruhota I think do you remember that our variables they are reals? – Michael Rozenberg Sep 02 '18 at 16:15
oh yes, some can be negative, however, how about max being less and $a>x_0$? – farruhota Sep 02 '18 at 16:18
@farruhota I proved the inequality in this case. See my solution. – Michael Rozenberg Sep 02 '18 at 16:19
1

oh, you are consider two cases: 1) max(a,b,c,d) is less than $x_0$; 2) $a>x_0$. And you are proving for each separately, OK, got it, thanks. +1 – farruhota Sep 02 '18 at 16:25

Elsa · Answer 4 · 2018-09-03T02:05:28.550

2

Alternatively, you could solve $min_{(a, b, c, d) |a+b+c+d=4} f(a, b, c, d),$ $f$ being your term with the squares. The convexity guarantees a global minimum. And the symmetry of the problem will yield the answer.

Edit According to the comments, $f$ is not convex but it is subharmonic (thanks to Jack D'Aurizio), i.e. for the given $f$ we have $\Delta f \geq 0$ on $\mathbb{R}^4$. $f$ is obviously not constant. So by the maximum principle it cannot have a maximum on the interior of its domain. Hence, it has a unique (unconstrained) minimum in $\mathbb{R}^4$. The constraint defining function $g(a, b, c, d) =a+b+c+d$ is harmonic. I did not go further but maybe it then can be shown that the constrained min problem has a unique solution and that it must be symmetric because otherwise we could move towards the origin (0,0,0,0) and further minimize the function (using a closed ball argument based in the definition of an harmonic function according to Wikipedia).

edited Sep 03 '18 at 02:05

answered Sep 02 '18 at 14:50

Elsa

948

@Elsa For which function did you use convexity? – Michael Rozenberg Sep 02 '18 at 14:57
@MichaelRozenberg $f=(a^2+3)(b^2+2)(c^2+3)(d^2+3)$ – Elsa Sep 02 '18 at 15:01
@Elsa Are you sure that it's a convex function? – Michael Rozenberg Sep 02 '18 at 15:04
@Severin Schraven Where do you see a solution? I think it's nothing. – Michael Rozenberg Sep 02 '18 at 15:20
I have to agree with Micheal. $f(a,b,c,d)$ is not even midpoint-convex. On the other hand $f$ is subharmonic, hence we may have some workaround. – Jack D'Aurizio Sep 02 '18 at 15:35

Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$

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