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# Problem from Inequality | PRMO-2018 | Problem 23

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

Answer:$$14$$

PRMO-2018, Problem 23

Pre College Mathematics

## Try with Hints

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 ..........

Therefore,

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

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

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

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

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

Therefore $$n=14$$

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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

Answer:$$14$$

PRMO-2018, Problem 23

Pre College Mathematics

## Try with Hints

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 ..........

Therefore,

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

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

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

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

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

Therefore $$n=14$$

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