←Vrati se na zadatke1. ZadatakDokaži da za sve pozitivne realne brojeve xxx i yyy vrijedi log2(xy)≥log(x2)log(y2).\log^2(xy)\ge \log(x^2)\log(y^2).log2(xy)≥log(x2)log(y2).Prikaži rješenjeRješenjeVrijedi log2(xy)=(logx+logy)2\log^2(xy)=(\log x+\log y)^2log2(xy)=(logx+logy)2 te log(x2)=2logx,log(y2)=2logy.\log(x^2)=2\log x,\qquad \log(y^2)=2\log y.log(x2)=2logx,log(y2)=2logy. Korištenjem tih formula dobivamo log2(xy)−log(x2)log(y2)=(logx+logy)2−4logxlogy=(logx−logy)2≥0.\begin{aligned} \log^2(xy)-\log(x^2)\log(y^2) &=(\log x+\log y)^2-4\log x\log y\\ &=(\log x-\log y)^2\ge0. \end{aligned}log2(xy)−log(x2)log(y2)=(logx+logy)2−4logxlogy=(logx−logy)2≥0.Pomakni formulu lijevo ili desno.