# How to solve this problem using Cauchy-Schwarz inequality

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.

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 $$k> 0$$.

• 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!
– user578340
Dec 13 '18 at 4:50
• I have added the two CS equivalent also ... Dec 13 '18 at 5:05