How to solve this problem? This is one of the problems in my semester examination:

For $x,y\in\mathbb{R}^+,x^2+y^2=1$, find the maximum value of $M=\sqrt{x}+\sqrt{2y}$.

I know it can be solved using the Cauchy-Schwarz inequality, but I can't solve it despite much effort with this idea.

Any help will be much appreciated! Although the examination is now over, but a solution for this problem will be very useful to me.

Thank you in advance.


1 Answer 1


Consider Hölder's inequality: $$(x^2+y^2)(1+2^{2/3})^3 \geqslant (\sqrt{x} + \sqrt{2y})^4$$

This can get equality when $x:y = 1:2^{1/3}$ which is certainly possible, so this determines the maximum of $\sqrt x + \sqrt{2y} $ as $(1+2^{2/3})^{3/4} \approx 2.04$.

P.S. Adding the CS inequality version as well: $$(x^2+y^2)(1+k) \geqslant (x+\sqrt k\, y)^2 \tag{$CS_1$}$$ $$(x+\sqrt k\, y)(1+k) \geqslant (\sqrt x+k^{3/4}\sqrt{y})^2 \tag{$CS_2$}$$

Now multiplying inequalities $(CS_1)$ and $(CS_2)^2$ together gives $$(x^2+y^2)(1+k)^3 \geqslant (\sqrt x+k^{3/4}\sqrt{y})^4$$ Setting $k=2^{2/3}$ makes the RHS what we want, which is the same as the Hölder's inequality above. You can work out that equality in both CS inequalities requires $x:y=1:\sqrt k$, which is possible for any $k> 0$.

  • $\begingroup$ Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that! $\endgroup$
    – user578340
    Dec 13, 2018 at 4:50
  • $\begingroup$ I have added the two CS equivalent also ... $\endgroup$
    – Macavity
    Dec 13, 2018 at 5:05

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