Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.

Question

Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$

I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$ Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$ which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.

So, how to solve the problem with elementary inequalities?

Answer 1

The solution by Vo Quoc Ba Can

Multiplying both sides by $a^2+b^2+c^2+ab+bc+ca,$ the inequality is equivalent to $$\sum {\frac{a^2+b^2+c^2+ab+bc+ca}{a^2+ab+b^2}} \geqslant \frac{9(a^2+b^2+c^2+ab+bc+ca)}{(a+b+c)^2},$$ equivalent to $$3+(a+b+c)\sum {\frac{c}{a^2+ab+b^2}} \geqslant 9-\frac{9(ab+bc+ca)}{(a+b+c)^2},$$ or $$(a+b+c)\sum {\frac{c}{a^2+ab+b^2}}+\frac{9(ab+bc+ca)}{(a+b+c)^2}\geq6.$$ By theCauchy-Schwartz inequality, we have $$\sum {\frac{c}{a^2+ab+b^2}}=\sum {\frac{c^2}{c(a^2+ab+b^2)}}\geq\frac{(a+b+c)^2}{\sum{c(a^2+ab+b^2)}}=\frac{a+b+c}{ab+bc+ca}.$$ Hence, it suffices to prove that $$\frac{(a+b+c)^2}{ab+bc+ca}+\frac{9(ab+bc+ca)}{(a+b+c)^2}\geq6.$$ Which is just AM-GM.

Answer 2

A more general inequality:

$$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 3(1+r)}{a^2+b^2+c^2+r(a+b+c)}$$ for $0\le r\le\frac{5}{2}$ is proved in Vasile Cirtoaje's "Algebraic Inequalities - Old and New Methods" using Equal-Value Theorem (Section 5.2, Problem 8, page 217).

Although for this case ($r=2$) Vo Quoc Ba Can's solution is much neater.

Answer 3

Another way.

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}\frac{1}{a^2+b^2+ab}-\frac{9}{(a+b+c)^2}=\sum_{cyc}\left(\frac{1}{a^2+b^2+ab}-\frac{3}{(a+b+c)^2}\right)=$$ $$=\frac{1}{(a+b+c)^2}\sum_{cyc}\frac{c^2-2a^2-2b^2-ab+2ac+2bc}{a^2+b^2+ab}=$$ $$=\frac{1}{2(a+b+c)^2}\sum_{cyc}\frac{(c-a)(c+b+4a)-(b-c)(a+c+4b)}{a^2+b^2+ab}=$$ $$=\frac{1}{2(a+b+c)^2}\sum_{cyc}(a-b)\left(\frac{a+c+4b}{b^2+c^2+bc}-\frac{c+b+4a}{a^2+c^2+ac}\right)=$$ $$=\frac{\sum\limits_{cyc}(a-b)^2(a^2+b^2-2c^2+5ab+2ac+2bc)(a^2+b^2+ab)}{2(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq$$ $$\geq\frac{\sum\limits_{cyc}(a-b)^2c(a+b-c)(a^2+b^2+ab)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq\frac{(a-c)^2b(a+c-b)(a^2+c^2+ac)+(b-c)^2a(b+c-a)(b^2+c^2+bc)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq$$ $$\geq\frac{(b-c)^2b(a-b)(a^2+c^2+ac)+(b-c)^2a(b-a)(b^2+c^2+bc)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}=$$ $$=\frac{(b-c)^2(a-b)^2(ab-c^2)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq0.$$

Answer 4

Multiplying the numerators of the fractions by $a^2$, $b^2$ and $c^2$ gives a wrong inequality: $$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{a^2}{a^2(b^2+c^2+bc)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}$$ and we need to prove that: $$(a+b+c)^4\geq9\sum\limits_{cyc}(2a^2b^2+a^2bc),$$ which is wrong for $c\rightarrow0^+$ and $a=b=1$.

But the following C-S helps already: $$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{(3a+b+c)^2}{(3a+b+c)^2(b^2+c^2+bc)}\geq\frac{25(a+b+c)^2}{\sum\limits_{cyc}(3a+b+c)^2(b^2+c^2+bc)}$$ and it's enough to prove that: $$25(a+b+c)^4\geq9\sum\limits_{cyc}(3a+b+c)^2(b^2+c^2+bc)$$ or $$\sum_{cyc}(7a^4+19a^3b+19a^3c-48a^2b^2+3a^2bc)\geq0$$ or $$3\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(4a^4+22a^3b+22a^3c-48a^2b^2)\geq0,$$ which is true by Schur and Muirhead.

Answer 5

Here is a proof following Momo answer.

$$\dfrac{1}{a^2+b^2+ab}+\dfrac{1}{b^2+c^2+bc}+\dfrac{1}{a^2+c^2+ac}\ge \dfrac{10.5}{a^2+b^2+c^2+2.5(ab+bc+ca)}$$ where all variables are nonnegative or positive. This is the inequality with $r=2.5$ also if the inequality in Momo answer is true for $r\ge 0$ it is also true for every $r_0\in[0,r]$ (not hard to prove), and $r=2.5$ is the maximum possible trying $a=b=1$ and $c\to 0$.

Since the inequality is quite symmetric we can multiply out denominators and with Muirhead inequality the case $r=2$ is easy. For $r=2.5$ we obtain the numerator $$7(b^5a+a^5b+c^5b+b^5c+c^5a+a^5c-(b^4a^2+a^4b^2+c^4b^2+b^4c^2+c^4a^2+a^4c^2))+N $$ where $N$ can be written as $$(a^3-c^3)(a-c)(a^2+c^2+ca-b^2)+(b^3-c^3)(b-c)(b^2+c^2+cb-a^2)+(a^3-b^3)(a-b)(a^2+b^2+ba-c^2).$$ Assuming $b\ge a\ge c$ we see that the only possible negative term is $ a^2+c^2+ca-b^2$ but $(a^2+c^2+ca-b^2)+(b^2+c^2+bc-a^2)\ge 0$ and this completes the proof.

Answer 6

I give another idea.

WLOG, we may assume that $a \ge b \ge c$. It's easy to see that the original problem follows immediately by adding two following inequalities: \begin{align} & \frac{1}{a^2+ab+b^2}+\frac{1}{b^2+bc+c^2} \ge \frac{4}{ab+bc+2ca+2b^2} \ \ \ \ \ \ \ \ (1) \\ &\frac{4}{ab+bc+2ca+2b^2} + \frac{1}{c^2+ca+a^2} \ge \frac{9}{\left(a+b+c \right)^2} \ \ \ \ \ \ \ \ (2)\end{align} Indeed, by reducing and factoring, the first inequality is equivalent with: \begin{align} \left(c-a \right)^2 \left(ab+bc+2ca-b^2 \right) \ge 0 \end{align} which is true because $a \ge b \ge c $.
For the second one, it can be proved by applying Cauchy-Schwarz as follows: \begin{align} \frac{4}{ab+bc+2ca+2b^2} &+ \frac{1}{c^2+ca+a^2} \ge \frac{ \left(2+1 \right)^2}{a^2+2b^2+c^2+ab+bc+3ca}\\ &= \frac{9}{ \left(a+b+c \right)^2 -\left(a-b \right) \left(b-c \right)} \ge \frac{9}{ \left(a+b+c \right)^2}\end{align} Hence, our proof is completed.