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# Inequality 37(Murray Klamkin)

Problem:

For all non-negative real numbers $\displaystyle a,b,c$ with sum $\displaystyle 2$, prove that

$\displaystyle (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)\leq 3$.

Solution:

Assume without loss of generality that

$\displaystyle a\geq b\geq c$. Moreover, denote by $\displaystyle t,u$ the $\displaystyle \frac{a+b}{2},\frac{a-b}{2}$.

Then we get that

$\displaystyle a=t+u, b=t-u$. From the hypothesis we deduce also that $\displaystyle t\leq 1$.

Let us now transform the $\displaystyle 3$ factors of the inequality in terms of $\displaystyle t,u$.

Thus we have that:

$\displaystyle a^2+ab+b^2=(t+u)^2+(t+u)(t-u)+(t-u)^2=3t^2+u^2$ and $\displaystyle (b^2+bc+c^2)(c^2+ca+a^2)=(t^2+tc+c^2)^2-u^2(2tc-c^2-u^2+2t^2)$.

Define by $\displaystyle f\left(a,b,c\right)$ the Left Hand Side of the Inequality, that is

$\displaystyle f\left(a,b,c\right)=(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)$.

We will now prove that $\displaystyle f(t,t,c)-f(a,b,c)\geq 0$. Denote the Left Hand Side by $\displaystyle X$. We must prove that $\displaystyle X\geq 0$.

We know that

$\displaystyle X=u^2(5t^4+4t^3c-6t^2c^2-2tc^2-c^4-u^4-u^2(t-c)^2)$.

We claim that the second factor of $\displaystyle X$ is greater than zero. Indeed, this is true as since $\displaystyle t=\max\left\{c,u\right\}$ we get that:

$\displaystyle 2t\geq t+c\Longrightarrow 4t^2\geq (t+c)^2\wedge t\geq c$.

Multiplying these $\displaystyle 2$ Inequalities we have that $\displaystyle 4t^3c\geq c^2(t+c)^2$ or $\displaystyle 4t^3c\geq t^2c^2+2tc^2+c^4$.  Adding up the $\displaystyle 5t^2c^2$ we get that

$\displaystyle 4t^3c+5t^2c^2\geq 6t^2c^2+2tc^3+c^4$.

Thus we have prove that

$\displaystyle 5t^4+4t^3c-6t^2c^2-2tc^2-c^4\geq 5t^4-5t^2c^2=5t^2(t-c)(t+c)$.

But from the last inequality we deduce that

$\displaystyle 5t^2(t-c)(t+c)\geq 5(t-c)t^3\geq 2(t-c)^4\geq u^2(t-c)^2+u^4$

due to the maximized value of $\displaystyle t$.

This completes the first scale of the proof. Now we only need to prove that if

$\displaystyle 2t+c=2$ then $\displaystyle 3t^2(t^2+tc+c^2)\leq 3$,

which is obviously true since it is of the form $\displaystyle (1-t)(3t^2-3t+1)\geq 0$, Q.E.D.