Largest k for $\displaystyle (a+b+c)^5 \ge\frac{k}{\sqrt{5}}(ab+bc+ca)(a-b)(b-c)(c-a)$

Question

The following problem ate up my mind , If $a,b,c$ are distinct non-negative real numbers then find the largest value of the constant $k$ auch that the following inequality is satisfied,

$\displaystyle (a+b+c)^5 \ge \frac{k}{\sqrt{5}}(ab+bc+ca)(a-b)(b-c)(c-a)$

Answer 1

I think it's better to find a maximum value of $k$ for which an inequality $$(a+b+c)^5\geq k(ab+ac+bc)(a-b)(b-c)(c-a)$$ is true for all non-negatives $a$, $b$ and $c$. The answer is $25\sqrt5$.

Indeed, we can assume $k>0$ and $a\geq c\geq b$ and since

$(ab+ac+bc)(a-b)(a-c)(c-b)\leq(a-c)a^2c^2$ it's $(a-c)(a^2+ac+c^2-ab-bc)\geq0$,

it remains to find a maximal value of k, for which the inequality $$(a+c)^5\geq k(a-c)a^2c^2$$

is true for all non-negatives $a$ and $c$ such that $a\geq c$.

It's enough to take $a>c>0$.

Let $a=xc$.

Hence, $x>1$ and $k_{max}=\min\limits_{x>1}\frac{(x+1)^5}{x^2(x-1)}=25\sqrt5$.

The last statement is obvious by AM-GM.