 Try this beautiful Algebra problem from PRMO, 2018 based on Inequality.

## Problem from Inequality | PRMO | Problem-23

What is the largest positive integer n such that $\frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$$\geq n(a+b+c)$.

• $20$
• $24$
• $13$

### Key Concepts

algebra

Inequality

But try the problem first…

Answer:$14$

Source

PRMO-2018, Problem 23

Pre College Mathematics

## Try with Hints

First hint

we have to find out the largest value of $n$…….

we know that ,

$\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z} \geq \frac{(a+b+c)^2}{x+y+z}$ .we may use this form to getting the largest positive integer $n$

Can you now finish the problem ……….

Second Hint

Therefore,

$\frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$$\geq \frac{(a+b+c)^2}{a(\frac{1}{29}+\frac{1}{31}) +b(\frac{1}{29}+\frac{1}{31})+c(\frac{1}{29}+\frac{1}{31})}$

$\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$ $\geq \frac{(a+b+c)}{(\frac{1}{29} +\frac{1}{31})}$

$\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$$\geq \frac{a+b+c}{\frac{60}{29 \times 31}}$

$\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$$\geq \frac{29\times 31}{60}(a+b+c)$

$\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$$\geq 14.98(a+b+c)$

Final Step

Therefore $n=14$