$a^2b^2+a^2c^2+b^2c^2\leq 3$

Question

Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$ Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.

My idea:

I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.

Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$ I have to show that $$y^2-2xz\leq 3$$ I tried to prove it with the sign of trinom, but it doesn't work.

Answer 1

We'll prove that $$(a^2-a+1)(b^2-b+1)\geq\frac{a^2+b^2}{2}.$$ Indeed, let $a+b=2u$ and $ab=v^2$.

Thus, we need to prove that $$a^2b^2-ab(a+b)+a^2+b^2+ab-a-b+1\geq\frac{a^2+b^2}{2}$$ or $$v^4-2uv^2+2u^2-2u+1\geq0$$ or $$(v^2-u)^2+(u-1)^2\geq0,$$ which is obvious. Thus, $$1=\prod_{cyc}(a^2-a+1)=\sqrt{\prod_{cyc}(a^2-a+1)(b^2-b+1)}\geq\sqrt{\prod_{cyc}\frac{a^2+b^2}{2}}.$$ Id est, it's enough to prove that: $$\sqrt{\prod_{cyc}\frac{a^2+b^2}{2}}\geq\left(\sqrt[4]{\frac{a^2b^2+a^2c^2+b^2c^2}{3}}\right)^3.$$ Now, let $a^2=x$, $b^2=y$ and $c^2=z$.

Thus, we need to prove that $$27(x+y)^2(x+z)^2(y+z)^2\geq64(xy+xz+yz)^3.$$ Now, since $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)$$ it's $$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that $$(x+y+z)^2\geq3(xy+xz+yz)$$ or $$\sum_{cyc}(x-y)^2\geq0$$ and we are done!

Answer 2

Another idea is to notice that:

$$2(a^2-a+1)^2=a^4+1+(a-1)^4\geq a^4+1$$

So, using this and Cauchy-Schwarz:

$$16 = 16(a^2-a+1)^2(b^2-b+1)^2(c^2-c+1)^2 \geq 2(a^4+1)(b^4+1)(c^4+1)=$$

$$=(a^4+a^4+1+1)(b^4+c^4+b^4c^4+1)\geq (a^2b^2+a^2c^2+b^2c^2+1)^2$$

It follows that:

$$a^2b^2+b^2c^2+c^2a^2 \leq 3$$

with equality when $a=b=c=1$.

Answer 3

Remark: I just realize that my Fact 1 is the same at Atticus's solution (he gave an elegant solution). My Fact 1 is derived from the optimization theory (the method of Lagrange Multipliers, penalty functions, etc.). We want to minimize $f(a,b,c) = 3 - (a^2b^2 + b^2c^2 + c^2a^2)$ subject to $g(a,b,c) = (a^2-a+1)(b^2-b+1)(c^2-c+1)-1 = 0$. Let $L(a,b,c) = f(a,b,c) + \lambda g(a,b,c)$ be the Lagrangian. At the optimum $(a, b, c) = (1,1,1)$, from $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial a} = \frac{\partial L}{\partial a} = 0$, we get $\lambda = 4$. I use Mathematica Resolve command to see that $L(a, b, c)\ge 0$ is true surprisingly! Actually, I first tried penalty functions. Find $\mu > 0$ such that $F(a,b,c) = f(a,b,c) + \mu [g(a,b,c)]^2 \ge 0$. But I found it is complicated.

An alternative solution:

I found the following fact (the proof is given later):

Fact 1: For any real numbers $a, b, c$, it holds that $$a^2b^2 + b^2c^2 + c^2a^2 \le 3 + 4\cdot [(a^2-a+1)(b^2-b+1)(c^2-c+1)-1].$$

From Fact 1, the desired result follows.

$\phantom{2}$

Proof of Fact 1: $\mathrm{RHS} - \mathrm{LHS}$ can be expressed as SOS (Sum of Squares). Indeed, we have $$\mathrm{RHS} - \mathrm{LHS} = \frac{1}{6} z^T Q z$$ where $z = [1, a, b, c, ab, ca, bc, abc]^T$ and ($Q$ is positive semidefinite) \begin{align} Q = \left(\begin{array}{rrrrrrrr} 18 & -12 & -12 & -12 & 7 & 7 & 7 & -3\\ -12 & 24 & 5 & 5 & -12 & -12 & -3 & 5\\ -12 & 5 & 24 & 5 & -12 & -3 & -12 & 5\\ -12 & 5 & 5 & 24 & -3 & -12 & -12 & 5\\ 7 & -12 & -12 & -3 & 18 & 7 & 7 & -12\\ 7 & -12 & -3 & -12 & 7 & 18 & 7 & -12\\ 7 & -3 & -12 & -12 & 7 & 7 & 18 & -12\\ -3 & 5 & 5 & 5 & -12 & -12 & -12 & 24 \end{array}\right). \end{align} The desired result follows.