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Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.

Problem 1

Prove that ab≤(a+b)/2\sqrt{ab} \le (a + b)/2 where aa and bb are positive real numbers.

Hints

Hint #1

See if you can convert the given statement into something that’s true.

Solution

Open to see the solution; try solving it first or see the hints above! ab≤?a+b2need to prove thisab≤?(a+b2)2=a2+2ab+b24square both sides4ab≤?a2+2ab+b2multiply both sides by 40≤?a2−2ab+b2simplify0≤(a−b)2factor \newcommand{\mle}{\stackrel{?}{\le}} \begin{align*} \sqrt{ab} & \mle \frac{a+b}{2} & \quad \text{need to prove this} \\ ab & \mle \left( \frac{a+b}{2} \right)^2 = \frac{a^2 + 2ab + b^2}{4} & \quad \text{square both sides} \\ 4ab & \mle a^2 + 2ab + b^2 & \quad \text{multiply both sides by $4$} \\ 0 & \mle a^2 - 2ab + b^2 & \quad \text{simplify} \\ 0 & \le (a - b)^2 & \quad \text{factor} \\ \end{align*}

Since (a−b)2(a - b )^2 is always non-negative, we know that the last statement 0≤(a−b)20 \le (a - b)^2 is true, which makes the original statement in the problem as true, since we showed that they’re equivalent. ■\quad \blacksquare