Prove that $\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\geq\sqrt[3]{45(ab+ac+bc)}$

Question

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\sqrt[3]{a^2+4bc}+\sqrt[3]{b^2+4ac}+\sqrt[3]{c^2+4ab}\geq\sqrt[3]{45(ab+ac+bc)}$$

A big problem in this inequality there is around $(1,1,0)$.

I tried Holder: $$\left(\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\right)^3\sum_{cyc}(a^2+4bc)^3(ka+b+c)^4\geq\left(\sum\limits_{cyc}(a^2+4bc)(ka+b+c)\right)^4$$ Thus, it remains to prove that $$\left(\sum\limits_{cyc}(a^2+4bc)(ka+b+c)\right)^4\geq45(ab+ac+bc)\sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$ which is nothing for all $k\geq0$.

Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.

I have a proof of the following inequality.

Let $a$, $b$ and $c$ be non-negative numbers and $k=8\cos^340^{\circ}.$ Prove that: $$\sqrt[3]{a^2+kbc}+\sqrt[3]{b^2+kac}+\sqrt[3]{c^2+kab}\geq\sqrt[3]{9(1+k)(ab+ac+bc)},$$

but it not so comforts.

Thank you!

score 7 · Answer 1 · answered Jul 25 '19 at 12:07

7

Assume that $ab+bc+ca > 0$. Rewrite the inequality as $\sqrt[3]{u} + \sqrt[3]{v} + \sqrt[3]{w} \ge 3$ where $$u = \frac{27(a^2+4bc)}{45(ab+bc+ca)}, \ v = \frac{27(b^2+4ca)}{45(ab+bc+ca)}, \ w = \frac{27(c^2+4ab)}{45(ab+bc+ca)}.$$

We will use the fact that $$\sqrt[3]{x} \ge \frac{3x(5+4x)}{5x^2+20x+2}, \ x \ge 0$$ which follows from $$x - \Big( \frac{3x(5+4x)}{5x^2+20x+2}\Big)^3 = \frac{x(125x^2+272x+8)(x-1)^4}{(5x^2+20x+2)^3} \ge 0.$$

Using the fact above, it suffices to prove that $$\frac{3u(5+4u)}{5u^2+20u+2} + \frac{3v(5+4v)}{5v^2+20v+2} + \frac{3w(5+4w)}{5w^2+20w+2} \ge 3$$ or $f(a,b,c) \ge 0$ where $f(a,b,c)$ is a homogeneous polynomial.

We use the Buffalo Way. WLOG, assume that $c = \min(a,b,c)$. There are two possible cases:

1) $c \le b\le a$: Let $b = c + s, \ a = c+s+t; \ s, t\ge 0$. Note that $f(c+s+t, c+s, c)$ is a polynomial with non-negative coefficients. True.

2) $c \le a \le b$: Let $a = c + s, \ b = c + s+t; \ s, t \ge 0$. Note that $f(c+s, c+s+t, c)$ is a polynomial with non-negative coefficients. True.

We are done.

answered Jul 25 '19 at 12:07

River Li

51,793

I checked a first inequality. It's true. I can not check the second. I have no an enough software. Thank you very much! – Michael Rozenberg Jul 25 '19 at 13:24
2

Hope you find some appropriate math software soon. – River Li Jul 25 '19 at 14:33
@dandide One of the methods for inequalities is to use bounds. So you find and check bounds if they are enough. – River Li Nov 23 '19 at 01:08
@RiverLi not a trivial bound I'm amazed by it, thanks again – Dabed Nov 23 '19 at 01:18
@dandide It is Pade approximation. – River Li Nov 23 '19 at 01:24
I tried PadeApproximant((x^(1/3)), {x,y, {2, 2}}) and I get $\frac{y^{\frac{1}{3}} (14 x^2 + 35 x y + 5 y^2)}{5 x^2 + 35 x y + 14 y^2}$ so no y gives me the same polynomial, could you help me get what I'm doing wrong? – Dabed Nov 23 '19 at 02:50
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$x^{1/3} = \frac{x}{x^{2/3}}$. PadeApproximant[[x^(2/3)], {x,1, {2, 1}}] – River Li Nov 23 '19 at 03:05
thank you very much – Dabed Nov 23 '19 at 04:21
Why did you choose that specific polynomial @River Li? – João Víctor Melo Feb 26 '21 at 15:56
@JoãoVíctorMelo Do you mean bounds? It is Pade approximation. – River Li Feb 26 '21 at 16:15
I mean "We will use the fact that..." why to use this fact? – João Víctor Melo Feb 26 '21 at 17:07
@JoãoVíctorMelo The purpose is to eliminate the root sign. The pade approximation is one of the approach to do this. – River Li Feb 26 '21 at 23:33

River Li · Answer 2 · 2025-09-09T04:17:47.977

My second solution.

WLOG, assume that $ab + bc + ca = 3$. We need to prove that $$\sqrt[3]{\frac{a^2+4bc}5}+\sqrt[3]{\frac{b^2+4ac}5}+\sqrt[3]{\frac{c^2+4ab}5} \ge 3.$$

Fact 1. Let $x, y, z\ge 0$ such that $xyz \ge 4/5$ and $$(x^3y^3 + y^3z^3 + z^3x^3)(x^3+y^3+z^3) + 18x^3y^3z^3 \ge 27.$$ Then $x + y + z \ge 3$.
(It is verified by Mathematica - a Computer Algebra System (CAS).)

$\phantom{2}$

Now, let $$x := \sqrt[3]{\frac{a^2+4bc}5}, \quad y := \sqrt[3]{\frac{b^2+4ac}5}, \quad z := \sqrt[3]{\frac{c^2+4ab}5}.$$

We can prove that $xyz \ge \frac45$ or $$(a^2+4bc)(b^2+4ca)(c^2+4ab) \ge 64.$$

Also, we can prove that $$(x^3y^3 + y^3z^3 + z^3x^3)(x^3+y^3+z^3) + 18x^3y^3z^3 \ge 27. \tag{1}$$ (Note: In pqr notation, $p = a + b + c, q = ab + bc + ca, r = abc$, (1) is written as $4\,{p}^{4}q+298\,{p}^{3}r+{p}^{2}{q}^{2}-1060\,pqr+58\,{q}^{3}+2250\,{ r}^{2}-3375 \ge 0.$ Perhaps the pqr method works.)

All the inequalities here are verified by Mathematica - a Computer Algebra System (CAS).

By Fact 1, the desired inequality follows.

Prove that $\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\geq\sqrt[3]{45(ab+ac+bc)}$

2 Answers2