If the minimum value of $x^{2}+x+1$ is $\frac{3}{4}$ . Find the maximum value of $\frac{x^{2}+x+4}{x^{2}+x+1}.$

This is a part of 10-seconds challenge .

The answer is 5.00.

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We can write $\frac{x^{2}+x+4}{x^{2}+x+1}=\frac{x^{2}+x+1+3}{x^{2}+x+1}=1+\frac{3}{x^{2}+x+1}$

For $\frac{x^{2}+x+4}{x^{2}+x+1}$ to be maximum $\implies$ $x^{2}+x+1$ to be minimum.

So maximum value of $\frac{x^{2}+x+4}{x^{2}+x+1}= 1+\frac{3}{\frac{3}{4}}=1+4=\boxed{5}$

This is 10 sec method.