Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.

Question

Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$

We have that $$a^2 + 4b^2 + c^2 = a^2 + (a + b + c + 1)b^2 + c^2 = (b^2 + a)a + (b + 1)b^2 + (b^2 + c)c$$

$$\implies \sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} = \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\frac{(a + b + c)^2}{(b^2 + a)a + (b + 1)b^2 + (b^2 + c)c}$$

$$ \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left(\frac{a}{b^2 + a} + \frac{1}{b + 1} + \frac{c}{b^2 + c}\right)$$

Furthermore, $\dfrac{a}{c^2 + a} + \dfrac{c}{a^2 + c} = 1 - \dfrac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}$, $$\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left[\frac{1}{b + 1} - \frac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}\right] + \frac{1}{3}$$

Then I don't know what to do next.

Answer 1

The idea is indeed that of using Cauchy-Schwarz. The problem is you went from a homogeneous expression ($a^2+4b^2+c^2$) to non-homogeneous terms and that may be harder to prove than the original inequality.

Instead, I would apply Cauchy-Schwarz like this:

$$\frac{(a+b+c)^2}{a^2+4b^2+c^2} \leq \frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{b^2+c^2}=\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}+\frac{1}{2}$$

and summing cyclically:

$$ \begin{aligned} \sum \frac{(a+b+c)^2}{a^2+4b^2+c^2} &\leq \sum\left(\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}\right)+\frac{3}{2}\\ &=\sum\left(\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}\right)+\frac{3}{2}\\ &=3+\frac{3}{2}=\frac{9}{2} \end{aligned} $$

Answer 2

We need to prove that $$\sum_{cyc}\frac{1}{b^2+c^2+4a^2}\leq\frac{9}{2(a+b+c)^2}$$ or $$\sum_{cyc}(2a^6-4a^5b-4a^5c+13a^4b^2+13a^4c^2-4a^4bc-12a^3b^3-12a^3b^2c-12a^3c^2b+20a^2b^2c^2)\geq0$$ or $$\sum_{cyc}(a-b)^2(2c^4+2(a^2-4ab+b^2)c^2+a^4-2a^3b+4a^2b^2-2ab^3+b^4)\geq0$$ for which it's enough to prove that: $$(a^2-4ab+b^2)^2-2(a^4-2a^3b+4a^2b^2-2ab^3+b^4)\leq0$$ or $$(a-b)^2(a^2+6ab+b^2)\geq0$$ and we are done!

Answer 3

Another way.

We'll prove that our inequality is true for any real $a$, $b$ and $c$ such that $a+b+c=3.$

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v^2$ can be negative, and $abc=w^3$.

Thus, by my first proof we need to prove that: $$\sum_{sym}(a^6-4a^5b+13a^4b^2-2a^4bc-6a^3b^3-12a^3b^2c+10a^2b^2c^2)\geq0$$ or $$27w^6+A(u,v^2)w^3+B(u,v^2)\geq0,$$ where $A$ and $B$ are polynomials of $u$ and $v^2$.

We'll prove that even the following inequality is true. $$\sum_{sym}(a^6-4a^5b+13a^4b^2-2a^4bc-6a^3b^3-12a^3b^2c+10a^2b^2c^2)\geq$$ $$\geq\frac{1}{3}\left(\sum_{cyc}(a^3-a^2b-a^2c+abc)\right)^2.$$ Since $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=27u^3-27uv^2+3w^3-9uv^2+3w^3+3w^3$$ and $$27-\frac{1}{3}\cdot9^2=0,$$ we see that the last inequality is a linear inequality of $w^3$.

Thus, it's enough to prove the last inequality for extreme value of $w^3$,

which happens for equality case of two variables.

Since the last inequality is homogeneous, it's enough to assume $b=c=1$, which gives $$2(a-1)^4(a^2+5)\geq\frac{1}{3}(a-1)^4(a+2)^2$$ or $$(a-1)^4(5a^2-4a+26)\geq0,$$ which is obvious.