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.

Answer 1

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).

Answer 2

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!

Answer 3

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.

Answer 4

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}