An algebraic inequality involving $\sum_{cyc} \frac1{(a+2b+3c)^2}$

Question

I was reading through the proof of an inequality posted on a different website and the following was mentioned as being easily proven by AM-GM:

Let $a,\ b,\ c>0$, then $$\frac{1}{(a+2b+3c)^2} + \frac{1}{(b+2c+3a)^2} + \frac{1}{(c+2a+3b)^2} \le \frac{1}{4(ab + bc+ca)}$$

I checked that the inequality holds numerically, but I can't find a solution (definitely nothing obvious using AM-GM). I played around with a simpler version of the inequality: $$\frac{1}{(a+2b)^2} + \frac{1}{(b+2a)^2} \le \frac{2}{9}\frac{1}{ab}$$ which I can prove by brute force algebra and a little AM-GM. However, brute force seems impractical for the three variable inequality. Any suggestions?

Answer 1

Hint: Using AM-GM, $$(a+c+2(b+c))^2=(a+c)^2+4(a+c)(b+c)+4(b+c)^2\geqslant 6(a+c)(b+c)+3(b+c)^2=3(b+c)(2a+b+3c)$$

Hence it is enough to show: $$\sum_{cyc} \frac{ab+bc+ca}{(b+c)(2a+b+3c)} \leqslant \frac34$$ $$\iff \sum_{cyc}\frac{(b+c)^2+2c^2}{(b+c)(2a+b+3c)}\geqslant\frac32$$ which follows almost directly from CS inequality, as its:

$$LHS \geqslant \frac{\left(\sum_{cyc} (b+c) \right)^2+2(\sum_{cyc} c)^2 }{\sum_{cyc} (b+c)(2a+b+3c)} = \frac{4(a+b+c)^2+2(a+b+c)^2 }{4(a+b+c)^2}=\frac32$$

Answer 2

A full expanding gives $$\sum_{cyc}(36a^6+80a^5b+104a^5c+21a^4b^2+189a^4c^2+58a^3b^3+82a^4bc-266a^3b^2c-122a^3c^2b-182a^2b^2c^2)\geq0,$$ which is true by AM-GM.

Also, SOS helps.

Let $2a+b=3x$, $2b+c=3y$ and $2c+a=3z$.

Thus, $a=\frac{x-2z+4y}{3}$, $b=\frac{y-2x+4z}{3}$, $c=\frac{z-2y+4x}{3}$, $$ab+ac+bc=\frac{1}{9}\sum_{cyc}(x-2z+4y)(y-2x+4z)=\frac{1}{3}\sum_{cyc}(5xy-2x^2)$$ and we need to prove that $$\sum_{cyc}\frac{1}{(x+y)^2}\leq\frac{27}{4\sum\limits_{cyc}(5xy-2x^2)}$$ or $$\sum_{cyc}\left(\frac{9}{4\sum\limits_{cyc}(5xy-2x^2)}-\frac{1}{(x+y)^2}\right)\geq0$$ or $$\sum_{cyc}\frac{17x^2+17y^2+8z^2-2xy-20xz-20yz}{(x+y)^2}\geq0$$ or $$\sum_{cyc}\frac{(y-z)(-x+17y-4z)-(z-x)(-y+17x-4z)}{(x+y)^2}\geq0$$ or $$\sum_{cyc}(x-y)\left(\frac{-z+17x-4y}{(x+z)^2}-\frac{-z+17y-4x}{(y+z)^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2(4x^2+4y^2+23z^2-13xy+9xz+9yz)}{(x+z)^2(y+z)^2}\geq0$$ or $$\sum_{cyc}(x^2-y^2)^2(4x^2+4y^2+23z^2-13xy+9xz+9yz)\geq0$$ or

$$\sum_{cyc}(x^2-y^2)^2(4x^2+4y^2-4xy+23z^2+9xz+9yz-9xy)\geq0,$$ for which it's enough to prove that $$\sum_{cyc}(x^2-y^2)^2(xz+yz-xy)\geq0.$$ Indeed, since the last inequality is symmetric, we can assume $x\geq y\geq z$.

Thus, $$\sum_{cyc}(x^2-y^2)^2(xz+yz-xy)\geq(x^2-y^2)^2(xz+yz-xy)+(x^2-z^2)^2(xy+yz-xz)\geq$$ $$\geq(x^2-y^2)^2(xz+yz-xy)+(x^2-y^2)^2(xy+yz-xz)=2yz(x^2-y^2)^2\geq0$$ and we are done!