Post #3142164

You wanted the proof of \[(m+q)(n+q)=2mn+q^2\] \[(m+q)(n+q)<2mn+\frac{mn}{k+2}\] \[(m+q)(n+q)<mn(2+\frac{1}{3})\] \[\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)<\frac{7}{3}\] \[2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)<\frac{14}{3}\] \[\left\{\left(\frac{m+q}{m}\right)+ \left(\frac{n+q}{n}\right)\right\}^2=3^2\] \[\left(\frac{m+q}{m}\right)^2+ 2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)+\left(\frac{n+q}{n}\right)^2=3^2\] \[9-\left\{\left(\frac{m+q}{m}\right)^2+ \left(\frac{n+q}{n}\right)^2\right\}=2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)<\frac{14}{3}\] \[9-\frac{14}{3}<\left(\frac{m+q}{m}\right)^2+ \left(\frac{n+q}{n}\right)^2\] \[\frac{13}{3}<(2-(k\epsilon^{\prime}+\epsilon)^2+(1+(k\epsilon^{\prime}+\epsilon)^2\] \[\frac{13}{3}-\frac{9}{4}<\left(\frac{m+q}{m}\right)^2\qquad \because 1+(k\epsilon^{\prime}+\epsilon)<\frac{3}{2}\] \[\therefore\sqrt{\frac{25}{12}}<q\approx \frac{14}{9}<\left(\frac{m+q}{m}\right)\] \[\because \frac{n}{k+2}\approx q\approx \frac{14}{9}\implies n\leqslant\frac{14(k+2)}{9}\] \[In fact \quad n<\frac{14(k+2)}{9}\]