Here are some inequalities I solved today/yesterday.

Problem (USAMTS): Find the ordered pair of real numbers \((x,y)\) that satisfies the equation below, and demonstrate that it is unique: \[\frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}.\]

Solution: The equation rearranges as: \[\frac{36}{\sqrt{x}}+9\sqrt{x}+\frac{9}{\sqrt{y}}+\sqrt{y}=42.\] Next, observe that by AM-GM: \[\frac{36}{\sqrt{x}}+9\sqrt{x}\geq2\sqrt{36\cdot9}=36,\] and: \[\frac{9}{\sqrt{y}}+\sqrt{y}\geq2\sqrt{9}=6.\] Adding these two inequalities yields: \[\frac{36}{\sqrt{x}}+9\sqrt{x}+\frac{9}{\sqrt{y}}+\sqrt{y}\geq42,\] so our equation is actually just the equality condition of AM-GM applied to \(x\) and \(y\) terms separately. Equality in AM-GM holds only when the terms are equal, hence the solution to the equation is unique. We have: \[\frac{36}{\sqrt{x}}=9\sqrt{x}\Rightarrow\boxed{x=\frac{36}{9}},\] and: \[\frac{9}{\sqrt{y}}=\sqrt{y}\Rightarrow\boxed{y=9}.\]

Problem: Show that for all positive real numbers \(x\neq1\) and nonnegative integers \(n\), we have \[\frac{x^{2n+1}-1}{x^{n+1}-x^n}\geq2n+1.\]

Solution: Observe that the LHS can be rewritten: \[\begin{split} \frac{x^{2n+1}-1}{x^{n+1}-x^n}&=\frac{x^{2n+1}-1}{x^n(x-1)}
&=\frac{x^{n+1}-\frac{1}{x^n}}{x-1}
&=\frac{\frac{1}{x^n}(1-x^{2n+1})}{1-x}. \end{split}\] This is the sum of a finite geometric series with first term \(\frac{1}{x^n}\), common ratio \(x\), and \(2n+1\) terms. Therefore, our inequality can be written as \[x^{-n}+x^{-n+1}+…+x^n\geq2n+1.\] But this is true by AM-GM on the LHS.

Since all of our steps are reversible, we are done. \(\square\)

Problem: Show that for all positive integers \(n\) with \(n\geq2\), we have \[\frac{1}{n}+\frac{1}{n+1}+…+\frac{1}{2n-1}>n(2^{1/n}-1).\]

Solution: We rearrange the inequality to: \[\frac{\frac{1}{n}+\frac{1}{n+1}+…+\frac{1}{2n-1}}{n}+1>2^{1/n}.\] This is simply: \[\frac{n+\frac{1}{n}+\frac{1}{n+1}+…+\frac{1}{2n-1}}{n}>2^{1/n}.\] Now, we break up the \(n\) in the numerator into \(n\) ones and allocate the ones to every remaining term in the numerator: \[\frac{\left(1+\frac{1}{n}\right)+\left(1+\frac{1}{n+1}\right)+…+\left(1+\frac{1}{2n-1}\right)}{n}>2^{1/n}.\] But now, this becomes: \[\frac{\frac{n+1}{n}+\frac{n+2}{n+1}+…+\frac{2n}{2n-1}}{n}>2^{1/n},\] which is true by AM-GM. Observe that the inequality is strict since obviously none of the terms in the numerator are equal (which is the equality condition for AM-GM).

Since all our steps are reversible, we are done. \(\square\)

I didn’t find (or solve) any interesting Cauchy-Schwarz problems. :(. Tomorrow/today I think I will work on combo.