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Find the minimum value of $u$ where

$x^2+y^2=1\;$ and $\;u =\displaystyle{\dfrac{ax^2+by^2}{\sqrt {a^2x^2+b^2y^2}}}$

Angelo
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Kiran Kiran
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Martin R Nov 14 '22 at 19:58

1 Answers1

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HINT: Let $$E=a^2x^2+b^2y^2$$

$$(a+b)\left(a x^2+b y^2\right)=E+a b\left(x^2+y^2\right)=E+ab$$ $$\Rightarrow \frac{a x^2+b y^2}{\sqrt{E}}=\frac{\sqrt{E}}{a+b}+\frac{\left(\frac{a b}{a+b}\right)}{\sqrt{E}}$$

Ekaveera Gouribhatla
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