Mathematical Enigma - VII

$a^{2} + b^{2} + c^{2}\geqslant ab + bc +ca$

$\Rightarrow (a+b+c)^{2} \geqslant 3( ab + bc + ca)$

$\Rightarrow 1 \geqslant 3( ab + bc + ca)$

$\Rightarrow \frac{1}{3} \geqslant ab + bc +ca$

Therefore, the maximum value of the expression is $\frac{1}{3}$ .

We have $a+b+c=1$ . $\implies (a+b+c)^{2}=1 \\ \implies a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=1 \\ \implies a^{2}+b^{2}+c^{2}-ab-bc-ca+3ab+3bc+3ca=1 \\ \implies \dfrac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]+3(ab+bc+ca)=1$

$3(ab+bc+ca)$ will be maximum when $\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$ is minimum that is: $\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]=0$ . $\therefore \text {max} (ab+bc+ca=\frac {1}{3})$

The problem is symmetrical in all three variables. The boundary, where one or more of the variables are zero, gives low value of the sum of products. So the maximum is where the variables are the largest possible. The only symmetrical solution for that is $a=b=c=\frac{1}{3}$ . This gives the sum of the products as $\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1}{3}$ .