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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.


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