Inequality - Maximize k such that

Question

I'm not sure how to solve this problem I came across online. Here it is:

Let $a, b, c$ be nonnegative reals. Maximize $k$ such that $$(a+b+c)\left(\dfrac 1a + \dfrac 1b + \dfrac 1c\right) + \dfrac{k(ab + bc+ca)}{a^2+b^2+c^2} \geq 9 + k.$$

What I know is this:

$a, b, c$ must be positive, because otherwise we'd have division by zero. I also know that $(a+b+c)\left(\dfrac 1a + \dfrac 1b + \dfrac 1c\right) \geq 9$ for positive $a,b,c$.

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a decreasing function.

Thus, it remains to prove our inequality for a maximal value of $w^3$.

We know that $a$, $b$ and $c$ are positive roots of the following equation $$(x-a)(x-b)(x-c)=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ or $$x^3-3ux^2+3v^2x=w^3,$$ which says that a graph of $f(x)=x^3-3ux^2+3v^2x$ and the line $y=w^3$

have three common points: $(a,f(a))$, $(b,f(b))$ and $(c,f(c))$.

Now, let $a\leq b\leq c$ and since $u^2\geq v^2$, we see that $$f'(x)=3x^2-6xu+3v^2=3(x^2-2ux+v^2)=3(x-x_1)(x-x_2).$$

Let $x_1\leq x_2$.

We see that $x_1>0$ and $\left(x_1,f(x_1)\right)$ is a maximum point and $\left(x_2,f(x_2)\right)$ is a minimum point.

Now, let $w^3$ increases and $u$ and $v^2$ are constants.

Thus, $a$, $b$ and $c$ are change and $w^3$ will get a maximal value,

when the line $y=w^3$ will touch to the graph of $f$,

which happens for equality case of two variables (for $a=b$ in our case).

Since our inequality is homogeneous, we can assume $b=c=1$, which gives $$(a+2)\left(\frac{1}{a}+2\right)-9\geq k\left(1-\frac{2a+1}{a^2+2}\right)$$ or $$\frac{2(a^2+2)}{a}\geq k.$$ Now, by AM-GM $$\frac{2(a^2+2)}{a}=2\left(a+\frac{2}{a}\right)\geq2\cdot2\sqrt2=4\sqrt2,$$ which gives the answer $4\sqrt2$.

Done!

Answer 2

Let $x,y,z$ are sides of tringle.

Then : $abc=pr^2,\ ab+bc+ca=r(r+4R),\ a+b=c=p $

So :

$$\dfrac{r+4R}{r}+k\cdot\dfrac{r(r+4R)}{p^2-2r(r+4R)} \ge 9+k$$

For fixed $R$ and $r$, minimum of left sides holdes for maximum $p$.

Maximum $p$ holdes for case $a=b$.

We can assume $a=b=1$:

$\Leftrightarrow \dfrac{2(2+c^2)}{c} \ge k \Rightarrow k\le 4\sqrt{2}$