Problem from Inequality | PRMO-2018 | Problem 23

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

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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