If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$

Question

Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that: $$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$

The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.

This inequality is a similar to the very many contest's inequalities, but nothing helps.

At least, I don't see how we can prove it.

An example of my trying.

We need to prove that $$\sum_{cyc}\frac{7-6a}{2+a^2}\geq1$$ or $$\sum_{cyc}\left(\frac{7-6a}{2+a^2}+1\right)\geq4$$ or $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4.$$

By C-S $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(2+a^2)(a+k)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(2+a^2)(a+k)^2}$$

Now we'll find a value of $k$, for which the equality in the last inequality occurs for $a=b=2$ and $c=\frac{1}{2}$.

Since in all equality case we have $$\frac{a-3}{(2+a^2)(a+k)}=\frac{b-3}{(2+b^2)(b+k)}=\frac{c-3}{(2+c^2)(a+k)},$$ we obtain: $$\frac{2-3}{(2+2^2)(2+k)}=\frac{\frac{1}{4}-3}{(2+\left(\frac{1}{4}\right)^2)(\frac{1}{4}+k)},$$ which gives $k=-\frac{9}{4}$.

Thus, it remains to prove that $$\left(\sum\limits_{cyc}(a-3)(4a-9)\right)^2\geq4\sum_{cyc}(2+a^2)(4a-9)^2,$$ which is wrong for $a=4$ and $b=c=\frac{1}{2}$.

Any hint?

Thank you!

Answer 1

Clearly, that we only need to prove this inequality for case:$\ a , b, c >0$

Let $a=e^{t_1}, b= e^{t_2}, c= e^{t_3} $

$$f(t_1)+f(t_2)+f(t_3) \ge1 \ , \ t_1+t_2+t_3=0$$

$$f(t)=\dfrac{7-6e^t}{2+e^{2t}}$$

$f'(t)=\dfrac{2e^t(e^{t}-3)(3e^t+2)}{(e^{2t}+2)^2}$

Minimum of $f(t)$ is attained at $t=\ln(3)>0 \Rightarrow \min\left( f(t_1)+f(t_2)+f(t_3)\right)$, for $t_1+t_2+t_3=0$ is attained only for case : $t_1,t_2,t_3 \le t_*$

Since $f''(t)=-\dfrac{2e^t(e^t-3+\sqrt{11})(e^t-3-\sqrt{11})(3e^t+2-\sqrt{22})(3e^t+2+\sqrt{22})}{9(e^{2t}+2)^3}$

we only need to consider the inequality in case : $t_1\le t_2=t_3 \le t_*$

$a={q^2} \ , \ b=c=\dfrac{1}{q}$

$\Leftrightarrow \dfrac{(3q^2+6q+5)(2q-1)^2(q-1)^2}{(2q^2+1)(q^4+2)}\ge 0$

Equality holdes for: $(a=b=c=1)\ $ and $\ (a=\dfrac{1}{4}, b=c=2)$

Answer 2

Michael Rozenberg's solution without UVW:

Michael Rozenberg gave a nice application of Cauchy-Bunyakovsky-Schwarz inequality. Then it suffices to prove that, for all $a, b, c > 0$ with $a b c = 1$, $$9(a + b + c - 1)^2 - 8(a^2 + b^2 + c^2) - 12 \ge 0.$$ Denote LHS by $f(a, b, c)$. WLOG, assume $c\ge 1$. We have $$f(a, b, c) - f(\sqrt{ab}, \sqrt{ab}, c) = (\sqrt{a} - \sqrt{b})^2(a + 2\sqrt{ab} + b + 18c - 18) \ge 0.$$ Also, we have $$f(\sqrt{ab}, \sqrt{ab}, c) = \frac{(\sqrt{c} + 5)(\sqrt{c} + 1)(\sqrt{c} - 2)^2(\sqrt{c} - 1)^2}{c} \ge 0.$$

We are done.