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### Prove that, $a(a+b)+b(b+1)\geq -\frac {1}{3}$

To prove that $a(a+b)+b(b+1)\geq -\frac {1}{3}$ is same as proving that $a(a+b)+b(b+1)+\frac {1}{3}\geq 0$.

Consider, $a(a+b)+b(b+1)+\frac {1}{3}$

Or, $b^2+b(a+1)+a^2+\frac {1}{3}$

The above can be considered as quadratic expression in b, whose coefficient of $b^2$ is 1.

Discriminant $\Delta$=$(a+1)^2-4.\left( {{a^2} + \frac{1}{3}} \right)$

= $-3a^2+2a-\frac {1}{3}$

= $- 3\left( {{a^2} - \frac{2}{3}a + \frac{1}{9}} \right)$

= $- 3{\left( {a - \frac{1}{3}} \right)^2}$

$\le 0$

Since coefficient of $b^2$ is 1 and discriminant is $\le 0$ for the quadratic expression in b, the quadratic expression itself is $\ge 0$.

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$