Prove hard inequality

Question

Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$

I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM-GM or Cauchy Schwarz alone.

FWIW holder's with $\sum_{cyc} (2a + b)$ twice yields the same failure as Titu's Lemma. — incertia, Oct 01 '19 at 04:35
I know this isn't exactly very helpful, but Mathematica reduces this to true so the inequality is indeed correct. — YiFan Tey, Oct 01 '19 at 04:53
Has anyone tried setting a homogeneous constraint and seen if that simplifies things? I've tried the standard $a + b + c = 1, ab + bc + ca = 1, abc = 1$ and i've made no progress along each of the simplifications. — incertia, Oct 01 '19 at 04:56
Bring the RHS to the LHS, and call the new LHS $f(a,b,c)$. It should be possible to show that all partial derivatives of $f$ vanish exactly when $a=b=c\neq0$, but I don't know how to do so. — YiFan Tey, Oct 01 '19 at 05:10
Possible duplicate of Prove $\sum{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ – found instantly with Approach0 — Martin R, Oct 01 '19 at 07:29

score 2 · Answer 1 · answered Oct 01 '19 at 05:26

2

After full expanding we need to prove that: $$\sum_{cyc}(4a^{5}b+4a^{5}c-12a^{4}b^{2}+12a^{4}c^{2}+5a^{3}b^{3}+8a^{4}bc-19a^{3}b^{2}c+5a^{3}c^{2}b-7a^{2}b^{2}c^{2})\geq0$$ or $$6\sum_{cyc}ab(a^{2}-b^{2}-2ab+2ac)^{2}+ \sum_{sym}(2a^{5}b-a^{3}b^{3}-4a^{4}bc+10a^{3}b^{2}c-7a^{2}b^{2}c^{2})\geq0.$$ Thus, it's enough to prove that: $$\sum_{cyc}(a^5b+a^5c-a^3b^3-4a^4bc+5a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0.$$ Can you end it now?

answered Oct 01 '19 at 05:26

Michael Rozenberg

194,933

This ends easily with AM-GM and Muirhead but the act of brutal expansion just feels wrong. – incertia Oct 01 '19 at 06:03
@incertia Check it. It's right! By the way, I proved the last inequality by $uvw$. – Michael Rozenberg Oct 01 '19 at 06:05
Yes I have verified that this indeed works. – incertia Oct 01 '19 at 06:07
If $uvw$ method is too foreign consider the symmetric inequality $(a - b)^2 (b - c)^2 (c - a)^2 \geq 0$ also of $6$-th degree. – incertia Oct 01 '19 at 07:12
@incertia I think it will get a proof, but by $uvw$ we'll get a solution immediately because the last inequality in my post is equivalent to $f(w^3)\geq0,$ where $f$ is a concave function. – Michael Rozenberg Oct 01 '19 at 07:15
The question is a duplicate, you already wrote two answers here: https://math.stackexchange.com/q/906972/42969. – Martin R Oct 01 '19 at 07:30
@incertia I am ready to show a full proof. If so, we can merge this topic with other topic. – Michael Rozenberg Oct 01 '19 at 08:00

score 1 · Answer 2 · answered Oct 01 '19 at 05:52

1

Let $x=2a+b,y=2b+c,z=2c+a$, then the inequality becomes:$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$$ By using $a^2+b^2+c^2\ge ab+bc+ca$, $$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{\left(xy\right)^2+\left(yz\right)^2+\left(zx\right)^2}{x^2y^2z^2}\ge\dfrac{x^2yz+xy^2z+xyz^2}{x^2y^2z^2}=\dfrac{xyz\left(x+y+z\right)}{x^2y^2z^2}=\dfrac{x+y+z}{xyz}$$ Then, by using $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}$, $$\dfrac{x+y+z}{xyz}=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{9}{xy+yz+zx}$$ Lastly, use $a^2+b^2+c^2\ge ab+bc+ca$ again, $$\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}\le\dfrac{27}{3xy+3yz+3zx}=\dfrac{9}{xy+yz+zx}$$ Therefore, $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$ is correct. Can you get it?

answered Oct 01 '19 at 05:52

MafPrivate

4,033

In your first step there is a big mistake. By the way, can up-voter explain us, why did you do it? – Michael Rozenberg Oct 01 '19 at 05:58
What's the problem? – MafPrivate Oct 01 '19 at 06:00
Check please a right side of your first inequality. – Michael Rozenberg Oct 01 '19 at 06:01
oops sorry i miscalculated – incertia Oct 01 '19 at 06:03

Prove hard inequality

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