2012 JBMO Problems/Problem 1

Section 1

Let $a,b,c$ be positive real numbers such that $a+b+c=1$ . Prove that $\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right ).$ When does equality hold?

Solution

The LHS rearranges to $\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6$ . Since $b+c=1-a$ we have that $\frac{b+c}{a}=\frac{1-a}{a}$ . Therefore, the LHS rearranges again to $\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6$ .

Now, distribute the $\sqrt{2}$ on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get $\frac{2-2a}{a} + \frac{2-2b}{b} + \frac{2-2c}{c}+12 \geq 4(\sqrt{\frac{2-2a}{a}} + \sqrt{\frac{2-2b}{b}} + \sqrt{\frac{2-2c}{c}})$ Let $\sqrt{\frac{2-2a}{a}} = A$ and similarly for $B$ and $C$ .

The inequality now simplifies to $A^2+B^2+C^2+12 \geq 4(A+B+C)$ Note that because $a$ , $b$ , and $c$ are positive real numbers less than $1$ , $A$ , $B$ , and $C$ are always positive real numbers. Rearranging terms shows that this further simplifies to $(A^2-4A+4)+(B^2-4A+4)+(C^2-4C+4)\geq0$ which equals $(A-2)^2+(B-2)^2+(C-2)^2\geq0$ By the trivial inequality we know that this is always true. Finally, we have equality when $A=B=C=2$ and $\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4$ Solving the equations yields that equality holds when $\boxed{a=b=c=\frac{1}{3}}$

Solution by Someonenumber011 :)

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2012 JBMO Problems/Problem 1

Section 1

Solution