Prove that: $a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4 $

Question

Let $a,b,c$ be real and $a+b+c=0.$ Prove that:$$a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4$$ I know we can make the following:
Let $bc\geq0$ thus we need to prove $$b^{12}+c^{12}+(b+c)^{12}\geq \frac{2049}{8} b^4c^4(b+c)^4$$or $$\frac{(2b+c)^2(b+2c)^2(b-c)^2(4b^6+12b^5c+99b^4c^2+178b^3c^3+99b^2c^4+12bc^5+4c^6)}{8}\geq 0\text{,}$$which is true.

But I'm trying to find another solution. Can be?

Answer 1

Let $ab\leq0$.

Thus, by AM-GM we obtain: $$a^{12}+b^{12}+c^{12}-\frac{2049}{8}a^4b^4c^4=$$ $$=3a^4b^4c^4+2(ab+ca+bc)^6-24a^2b^2c^2(ab+ac+bc)^3-\frac{2049}{8}a^4b^4c^4=$$ $$=\frac{1}{8}(16(ab+ac+bc)^6-192(ab+ac+bc)^3a^2b^2c^2-2025a^4b^4c^4)=$$ $$=\frac{1}{8}(4(ab+ac+bc)^3-75a^2b^2c^2)(4(ab+ac+bc)^3+27a^2b^2c^2)=$$ $$=\frac{1}{8}(4(a^2+ab+b^2)^3+75a^2b^2c^2)(4(a^2+ab+b^2)^3-27a^2b^2c^2)=$$ $$=\frac{1}{8}(4(a^2+ab+b^2)^3+75a^2b^2c^2)\left(4\left((a+b)^2+2\left(-\frac{ab}{2}\right)\right)^3-27a^2b^2c^2\right)\geq$$ $$\geq\frac{1}{8}(4(a^2+ab+b^2)^3+75a^2b^2c^2)\left(4\left(3\sqrt[3]{(a+b)^2\left(-\frac{ab}{2}\right)^2}\right)^3-27a^2b^2c^2\right)=0.$$

Answer 2

Alternative proof:

If $abc = 0$, the desired inequality is clearly true.

If $abc \ne 0$, WLOG, assume that $abc = -1$. WLOG, assume further that $a, b > 0$ and $c < 0$. We need to prove that $a^{12} + b^{12} + c^{12} \ge \frac{2049}{8}$.

Let $q = ab + bc + ca$. It is easy to obtain $a^{12} + b^{12} + c^{12} = 2q^6-24q^3+3$. The details are given at the end.

Thus, it suffices to prove that $$2q^6-24q^3+3 \ge \frac{2049}{8}$$ or (since $q < 0$) $$q^3 \le - 27/4.$$

We have $a^3 + qa + 1 = a^3 + (ab + bc + ca)a + 1 = a^2(a + b + c) + abc + 1 = 0$. Thus, we have $q = - a^2 - \frac{1}{a} = - (a^2 + \frac{1}{2a} + \frac{1}{2a}) \le -3 \sqrt[3]{a^2 \cdot \frac{1}{2a} \cdot \frac{1}{2a}} = - 3\sqrt[3]{1/4}$. Thus, we have $q^3 \le - 27/4$.

We are done.

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The details of $a^{12} + b^{12} + c^{12} = 2q^6-24q^3+3$:

From $a + b + c = 0$ and $abc = -1$, it is easy to verify that \begin{align*} a^3 + qa + 1 &= 0, \\ b^3 + qb + 1 &= 0, \\ c^3 + qc + 1 &= 0. \end{align*} We have \begin{align*} &a^2 + b^2 + c^2 = (a + b + c)^2 - 2q = -2q, \\ &a^3 + b^3 + c^3 = -q(a + b + c) - 3 = - 3, \\ &a^4 + b^4 + c^4 = -q(a^2 + b^2 + c^2) - (a + b + c) = 2q^2. \end{align*} Then, we have \begin{align*} &a^{12} + b^{12} + c^{12}\\ =\,& (-qa - 1)^4 + (-qb - 1)^4 + (-qc - 1)^4\\ =\, & q^4(a^4 + b^4 + c^4) + 4q^3(a^3 + b^3 + c^3) + 6q^2(a^2 + b^2 + c^2) + 4q(a + b + c) + 3\\ =\, & 2q^6-24q^3+3. \end{align*}

Answer 3

HINT.- (COMMENT to get another kind of proof). Consider the beam of curves $$x^{12}+y^{12}+a=bx^4y^4;\space\space a,b\ge0$$ which are easily seen to be symmetric respect to the diagonals $y=\pm x$ and the axis $y=0$ and $x=0$.When $a=0$ and $b\ne 0$ the curve has four foliums concurrent at origin each of them placed in distinct quadrants and when $ab\gt0$ these foliums are disjoints (see for example $x^{12}+y^{12}+1=7x^4y^4$ and $x^{12}+y^{12}=12x^4y^4$).

Now for $z=c$ fixed, each of the four possibilities $y=\pm x\pm c$ gives place to the same (plane) curve $\Gamma_c$ as is easily verified $$\Gamma_c:\space \space x^{12}+y^{12}+c^{12}=\frac{2049}{8}c^4x^4y^4$$ The only points involved in our problem of inequality for the curve $\Gamma_c$ should be in the line, say, $y=x-c$ and noting that $\dfrac{2049}{8}=\dfrac{2^{12}+2}{2^4}$ we can see without effort that the curve $$\Gamma_c:\space \space x^{12}+(x-c)^{12}+c^{12}=\frac{2049}{8}c^4x^4y^4$$ has the points $P=(2c,c),Q=(\frac c2,-\frac c2),R=(-c,-2c)$. We can see also that in the diagonal $y=x$ there is (by simple symmetry) the point $S=(\frac c2,\frac c2)$ and the not easy to calculate $T=(\sqrt[4]{6\sqrt{114}+64}\space c,\sqrt[4]{6\sqrt{114}+64}\space c)$. It follows by symmetry all the points in the attached figure:

All these points are related to the other possible lines $y=\pm x\pm c$ which give, as I said, the same curve but in the case we want to solve we are concerned only with the points in the (red) line $y=x-c$.

►For all the points inside the foliums (shading region) we have the opposite of the inequality and out of it its validity so if we prove that the line $y=x-c$ is tangent to the curve, we are done.

►We leave this end to the reader who wanted to do it for which I think at two possible ways: consider the definition of tangent or calculate that at neighborhood of the points $P,Q,R$ in the line $y=x-c$ the points distinct of them verify the proposed inequality.