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Inequality

If $$a$$ and $$b$$ are real numbers satisfying $$3a+4b=15$$, find the minimum value of $$\sqrt{a^2 + b^2}$$.

Note by Raj Mantri
7 months, 3 weeks ago

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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 $$\boxed{3}.$$ · 7 months, 1 week ago

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 · 7 months, 1 week ago