ISI Interview Question 2013

Algebra Level 4

If $a+b+c=7$ and $a,b,c, \in \mathbb R^{+}$ then the minimum value of $\sum_{cyc}^{a,b,c} \dfrac{a^2+b^2}{a+b}=?$

The answer is 7.0.

1 solution

Skye Rzym
Jun 22, 2017

Given that $a+b+c=7$ and $a,b,c \in \mathbb R^{+}$ . $\sum_{cyc}^{a,b,c} \frac{a^{2}+b^{2}}{a+b}=\sum_{cyc}^{a,b,c} \frac{(a+b)^{2}-2ab}{a+b}$ $=\sum_{cyc}^{a,b,c} a+b-\frac{2ab}{a+b}=\sum_{cyc}^{a,b,c} a+b-\frac{2}{\frac{1}{a}+\frac{1}{b}}$ $\geq\sum_{cyc}^{a,b,c} a+b-\frac{a+b}{2}=\sum_{cyc}^{a,b,c} \frac{a+b}{2}=a+b+c=7$ or $\sum_{cyc}^{a,b,c} \frac{a^{2}+b^{2}}{a+b}=\sum_{cyc}^{a,b,c} \frac{a^{2}}{a+b}+\frac{b^{2}}{a+b}$ $\geq \sum_{cyc}^{a,b,c} \frac{(a+b)^{2}}{2(a+b)}=\sum_{cyc}^{a,b,c} \frac{a+b}{2}=a+b+c=7$

Better ways are there. I mean, titu's lemma can be used for solving this easily

Md Zuhair - 3 years, 11 months ago

One can just use trivial inequalities $(a-b)^2>=0$ .So $(a+b)^2>=(1/2)(a+b)^2$ .Now I guess its clear what's to be done next..That the minimum value is $(a+b+c)$

Spandan Senapati - 3 years, 11 months ago

from where u got to know the interview questions @Md Zuhair

rajdeep brahma - 3 years ago

Senior told me.

Md Zuhair - 3 years ago

ISI Interview Question 2013

The answer is 7.0.

1 solution

0 pending reports