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, postmortes Jun 26 at 16:30

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  • 4
    $\begingroup$ If you showed your Lagrange Multiplier solution, your question might be reopened. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – J_P Jun 13 at 14:26
  • $\begingroup$ @J_P It was typo. I fixed. Thank you! $\endgroup$ – Michael Rozenberg Jun 13 at 14:27
  • $\begingroup$ Which inequality is this again? I looked up Hölder on Wikipedia but there's just a lot of stuff about measure spaces and the like. $\endgroup$ – J_P Jun 13 at 14:43
  • $\begingroup$ @J_P See here: math.stackexchange.com/tags/holder-inequality/info $\endgroup$ – Michael Rozenberg Jun 13 at 14:45
  • 1
    $\begingroup$ Wow, great! Thanks. $\endgroup$ – J_P Jun 13 at 14:46

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