For any real number a,b,x,y,we have inequality \(ax+by\leq \sqrt{a^2+b^2}\times\sqrt{x^2+y^2}\) by using Cauchy-Swartz inequality.Hence, \(15=3a+4b\leq\sqrt{3^2+4^2}\times\sqrt{a^2+b^2}=5\sqrt{a^2+b^2}.\)
Therefore,the minimum value of \(\sqrt{a^2+b^2}\) is \(\boxed{3}.\)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestFor any real number a,b,x,y,we have inequality \(ax+by\leq \sqrt{a^2+b^2}\times\sqrt{x^2+y^2}\) by using Cauchy-Swartz inequality.Hence, \(15=3a+4b\leq\sqrt{3^2+4^2}\times\sqrt{a^2+b^2}=5\sqrt{a^2+b^2}.\)

Therefore,the minimum value of \(\sqrt{a^2+b^2}\) is \(\boxed{3}.\)

Log in to reply

using cauchy-swarch inequality in 3a+4b we get what ayush rai got u can check what is cauchy swarck inequality if u dont know

Log in to reply

So you were checking out my profile.How was my solution for the geometry problem.upvote if u like.

Log in to reply