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 \(\boxed{3}.\)

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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 \(\boxed{3}.\)

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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

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

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