3

How can we prove that:

$$a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10}+a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20}\geq 2(a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}), \ \forall\ a,b,c\geq 0.$$

I proved only that $32S_1+18S_2\geq 2 S$, using AM-GM inequality, where

$S_1=a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10},$ $S_2=a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20},$ $S=a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}$

Note that Muirhead's inequality is not working there...

Bogdan
  • 1,867
  • Possibly that should be $32S_1+18S_2 \ge 50 S$, which is what I get when I do AM-GM with some of those terms. – Macavity Mar 14 '16 at 07:28
  • Note that – presuming the proposed inequality holds true – the range of validity is all of $,\mathbb R^3$ since the LHS exhibits even powers of $a,b$, and $c$ only, whence no impact by variables changing sign, whereas the RHS will not increase. Equivalently, the difference LHS $-$ RHS is a positive polynomial. – Hanno Oct 12 '17 at 12:53

1 Answers1

0

Update 2019/10/05 (Simplify some proof.)

With computer, here is a solution:

Clearly, we only need to prove the case when $a, b, c > 0$.

Let $x=a^{10}, \ y = b^{10}, \ z=c^{10}$. The inequality is written as $$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge 2x^5yz\sqrt[10]{\frac{x}{y}} + 2xy^5z\sqrt[10]{\frac{y}{z}} + 2xyz^5\sqrt[10]{\frac{z}{x}}.$$

We will use the following bound: $$\sqrt[10]{u} \le f(u) = \frac{11 u^2 + 418 u + 171}{20(19 u + 11)}, \quad \forall u > 0$$ which follows from (letting $u = v^{10}$) \begin{align} &11 v^{20} + 418 v^{10} + 171 - 20(19v^{10}+11)v\\ =\ & (v-1)^4\\ &\times \left(11 v^{16}+44 v^{15}+110 v^{14}+220 v^{13}+385 v^{12}+616 v^{11}+924 v^{10}+1320 v^9+1815 v^8\right.\\ &\qquad \left. +2040 v^7+2044 v^6+1876 v^5+1585 v^4+1220 v^3+830 v^2+464 v+171\right). \end{align}

With the bound above, it suffices to prove that $$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge 2x^5yzf(x/y) + 2xy^5zf(y/z) + 2xyz^5f(z/x)$$ or $F(x,y,z)\ge 0$ where $F(x,y,z)$ is a homogeneous polynomial of degree $10$.

We use the Buffalo Way. WLOG, assume that $z = \min(x,y,z) $. There are two possible cases:

1) $z \le y \le x$: Let $z=1, \ y=1+s, \ x = 1+s + t; \ s,t \ge 0$. Then, $F(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients.

2) $z \le x \le y$: Let $z=1, \ x=1+s, \ y=1+s+t; \ s,t\ge 0$. Then, $F(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients.

We are done.

River Li
  • 37,323