# Minimum of $x^2+y^2$ for $\frac1x+\frac8y=1$ [closed]

Consider positive reals $$x$$ and $$y$$ such that $$\frac{1}{x}+\frac{8}{y}=1$$.

What is minimum of $$x^2+y^2$$?

I can solve this using differentiation or Lagrange multipliers but can someone suggest more elegant solution (e.g., using Cauchy-Schwarz, AM-GM, etc.)?

## closed as off-topic by RRL, Xander Henderson, metamorphy, José Carlos Santos, postmortesJun 26 at 16:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Xander Henderson, metamorphy, José Carlos Santos, postmortes
If this question can be reworded to fit the rules in the help center, please edit the question.

• If you showed your Lagrange Multiplier solution, your question might be reopened. – robjohn Jul 1 at 7:22

By Holder $$x^2+y^2=(x^2+y^2)\left(\frac{1}{x}+\frac{8}{y}\right)^2\geq\left(\sqrt[3]{x^2\cdot\left(\frac{1}{x}\right)^2}+\sqrt[3]{y^2\cdot\left(\frac{8}{y}\right)^2}\right)^3=125.$$ The equality occurs for $$(x^2,y^2)||\left(\frac{1}{x},\frac{8}{y}\right)$$ or $$x^3=\frac{1}{8}y^3$$ or $$y=2x,$$ which says that we got a minimal value.
• Graphing $1/x + 8/y$, I believe that $x^2+y^2\geq 125$ with equality at $x=5, y=10$ so I'm not sure about this solution. – J_P Jun 13 at 14:26