What is the largest positive integer n such that (a^2)/((b)/(29)+(c)/(31))+(b^2)/((c)/(29)+(a)/(31))+(c^2)/((a)/(29)+(b)/(31)) ≥ n(a+b+c) holds for all positive real numbers a,b,c.

Correct answer: 14 Since (a^2)/(x)+(b^2)/(y)+(c^2)/(z) ≥ ((a+b+c)^2)/(x+y+z) So, (a^2)/((b)/(39)+(c)/(31))+(b^2)/((c)/(29)+(a)/(31))+(c^2)/((a)/(29)+(b)/(31)) ≥ ((a+b+c)^2)/(a((1)/(29)+(1)/(31))+b((1)/(29)+(1)/(31))+c((1)/(29)+(1)/(31))) ≥ ((a+b+c))/(((1)/(29)+(1)/(31))) ≥ (a+b+c)/((60)/(29 × 31)) ≥ (29 × 31)/(60)(a+b+c) ≥ 14.98(a+b+c) So, n=14

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

What is the largest positive integer n such that...

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)

holds for all positive real numbers a,b,c.

Correct answer

Step-by-step solution

Since

\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z} \geq \frac{(a+b+c)^2}{x+y+z}

So,

\frac{a^2}{\frac{b}{39}+\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\left(\frac{1}{29}+\frac{1}{31}\right)+b\left(\frac{1}{29}+\frac{1}{31}\right)+c\left(\frac{1}{29}+\frac{1}{31}\right)}

\geq \frac{(a+b+c)}{\left(\frac{1}{29}+\frac{1}{31}\right)} \geq \frac{a+b+c}{\frac{60}{29 \times 31}} \geq \frac{29 \times 31}{60}(a+b+c) \geq 14.98(a+b+c)

So,

\mathrm{n}=14

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