Prove $2(a^2+b^2+c^2)+\frac{4}{3}\sum\limits_{\mathrm{cyc}}\frac{1}{a^2+1} \geq 5$ for $ab+bc+ac=1$

Question

The positive real numbers a, b, and c satisfy ab+bc+ac=1. Prove that the inequality $$ 2 \left(a^2+b^2+c^2\right)+\dfrac{4}{3}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right) \geq 5 $$ is true.

My solution for the first part is: From the inequality $$x^2+y^2+z^2 \geq \dfrac{\left(x+y+z\right)^2}{3}$$ we get $$ 2 \left(a^2+b^2+c^2\right) \geq \dfrac{2}{3}\left(a+b+c\right)^2.$$ Then I use the substitution $p=a^2+b^2+c^2, q=ab+bc+ac$ and this formula $a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)$. So we get that $$\left(a+b+c\right)^2=p+2q=p+2.$$ So $$ 2 \left(a^2+b^2+c^2\right) \geq \dfrac{2}{3}\left(a+b+c\right)^2=\dfrac{2}{3}\left(p+2\right).$$

any idea how to finish with the second part of the inequality? Thanks in advance!

Answer 1

Another way.

We need to prove that: $$\frac{2(a^2+b^2+c^2)}{ab+ac+bc}+\frac{4}{3}(ab+ac+bc)\sum_{cyc}\frac{1}{a^2+ab+ac+bc}\geq5$$ or $$\frac{2(a^2+b^2+c^2)}{ab+ac+bc}+\frac{8(a+b+c)(ab+ac+bc)}{3(a+b)(a+c)(b+c)}\geq5$$ or $$\sum_{sym}(6a^4b-a^3b^2-a^3bc-4a^2b^2c)\geq0,$$ which is true by Muirhead.

We can show that the minimal value of $k$, for which the inequality $$k(a^2+b^2+c^2)+\dfrac{4}{3}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\geq k+3$$ is true for any non-negatives $a$, $b$ and $c$ such that $ab+ac+bc=1$, it's $k=\frac{1}{3}.$

Answer 2

We can end you idea by the following way.

After your first step we need to prove that: $$\frac{2(a+b+c)^2}{3(ab+ac+bc)}+\frac{4}{3}(ab+ac+bc)\sum_{cyc}\frac{1}{a^2+ab+ac+bc}\geq5$$ or $$\frac{2(a+b+c)^2}{3(ab+ac+bc)}+\frac{4}{3}(ab+ac+bc)\sum_{cyc}\frac{1}{(a+b)(a+c)}\geq5$$ or $$\frac{2(a+b+c)^2}{3(ab+ac+bc)}+\frac{\frac{4}{3}(ab+ac+bc)\sum\limits_{cyc}(b+c)}{\prod\limits_{cyc}(a+b)}\geq5$$ or $$\frac{2(a+b+c)^2\prod\limits_{cyc}(a+b)}{3(ab+ac+bc)}+\frac{8}{3}(a+b+c)(ab+ac+bc)\geq5\prod\limits_{cyc}(a+b)$$ and since $$\prod\limits_{cyc}(a+b)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's just $$\sum_{cyc}c(a-b)^2\geq0,$$ it's enough to prove that: $$\frac{2(a+b+c)^2\cdot\frac{8}{9}(a+b+c)(ab+ac+bc)}{3(ab+ac+bc)}+\frac{8}{3}(a+b+c)(ab+ac+bc)\geq5\prod\limits_{cyc}(a+b)$$ or $$16(a+b+c)^3+72(a+b+c)(ab+ac+bc)\geq135(a+b)(a+c)(b+c)$$ or $$\sum_{cyc}(16a^3-15a^2b-15a^2c+14abc)\geq0,$$ which is true by Schur: $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0$$ and Muirhead:$$\sum_{cyc}(a^2b+a^2c-2abc)\geq0.$$

Answer 3

By AM-GM twice, we have \begin{align*} &2 \left(a^2+b^2+c^2\right)+\frac{4}{3}\left(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\\ ={}& 2(a^2 + b^2 + c^2) + \sum_{\mathrm{cyc}} \left(\frac43\cdot \frac{1}{a^2 + 1} + \frac34(a^2 + 1)\right) - \sum_{\mathrm{cyc}} \frac34(a^2 + 1)\\ \ge{}& 2(a^2 + b^2 + c^2) + \sum_{\mathrm{cyc}} 2\sqrt{\frac43\cdot \frac{1}{a^2 + 1} \cdot \frac34(a^2 + 1)} - \sum_{\mathrm{cyc}} \frac34(a^2 + 1)\\ ={}& \frac54(a^2 + b^2 + c^2) + \frac{15}{4}\\ ={}& \frac{5}{8}(a^2 + b^2) + \frac58(b^2+c^2) + \frac58(c^2 + a^2) + \frac{15}{4} \\ \ge{}& \frac{5}{8}\cdot 2ab + \frac58\cdot 2bc + \frac58\cdot 2ca + \frac{15}{4} \\ ={}& \frac54(ab + bc + ca) + \frac{15}{4}\\ ={}& 5. \end{align*}

We are done.