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Question Number 71206 by naka3546 last updated on 13/Oct/19

$$Letp,q,rarepositiverealnumbers.$$ $$0<r<min\{p,q\}.$$ $$Provethat$$ $$\sqrt{p-r}+\sqrt{q-r}⩽min\{\sqrt{\frac{pq}{r}},\sqrt{2(p+q-2r)}\}$$

Answered by mind is power last updated on 13/Oct/19

$$\sqrt{\mathrm{p}-\mathrm{r}}+\sqrt{\mathrm{q}-\mathrm{r}}=⩽\sqrt{2(\mathrm{p}+\mathrm{q}-2\mathrm{r})}$$ $$\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}⩽\sqrt{2\mathrm{x}+2\mathrm{y}}$$ $$\mathrm{cause}\mathrm{x}+\mathrm{y}-2\sqrt{\mathrm{xy}}={(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})}^2⩾0$$ $$\Rightarrow \mathrm{x}+\mathrm{y}+2\sqrt{\mathrm{xy}}⩽2\mathrm{x}+2\mathrm{y}$$ $$\Rightarrow \sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}⩽\sqrt{2(\mathrm{x}+\mathrm{y})}$$ $$\Rightarrow \mathrm{x}=\mathrm{p}-\mathrm{r}\mathrm{y}=\mathrm{q}-\mathrm{r}$$ $$\sqrt{\mathrm{p}-\mathrm{r}}+\sqrt{\mathrm{q}-\mathrm{r}}⩽\sqrt{2(\mathrm{p}+\mathrm{q}-2\mathrm{r})}....1$$ $$\sqrt{\mathrm{p}-\mathrm{r}}+\sqrt{\mathrm{q}-\mathrm{r}}⩽\sqrt{\frac{\mathrm{pq}}{\mathrm{r}}}$$ $$\sqrt{\mathrm{p}-\mathrm{r}}+\sqrt{\mathrm{q}-\mathrm{r}}=\sqrt{\mathrm{r}}\sqrt{\frac{\mathrm{p}}{\mathrm{r}}-1}+\sqrt{\mathrm{r}}\sqrt{\frac{\mathrm{q}}{\mathrm{r}}-1}$$ $$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{y}-1}⩽\sqrt{\mathrm{xy}}$$ $$\mathrm{x}+\mathrm{y}-2+2\sqrt{(\mathrm{x}-1)(\mathrm{y}-1)}⩽\mathrm{xy}$$ $$2\sqrt{(\mathrm{x}-1)(\mathrm{y}-1)}⩽\mathrm{xy}-\mathrm{x}-\mathrm{y}+2=(\mathrm{x}-1)(\mathrm{y}-1)+1$$ $$\Rightarrow {(\sqrt{(\mathrm{x}-1)(\mathrm{y}-1)}-1)}^2⩾0\mathrm{True}$$ $$\mathrm{x}=\frac{\mathrm{p}}{\mathrm{r}},\mathrm{y}=\frac{\mathrm{q}}{\mathrm{r}}\Rightarrow$$ $$\sqrt{\frac{\mathrm{p}}{\mathrm{r}}-1}+\sqrt{\frac{\mathrm{q}}{\mathrm{r}}-1}⩽\sqrt{\frac{\mathrm{pq}}{{\mathrm{r}}^2}}$$ $$\Rightarrow \sqrt{\mathrm{r}}(\sqrt{\frac{\mathrm{p}}{\mathrm{r}}-1}+\sqrt{\frac{\mathrm{q}}{\mathrm{r}}-1})⩽\sqrt{\frac{\mathrm{pq}}{\mathrm{r}}}...2$$ $$2\mathrm{\&}1\Rightarrow$$ $$\sqrt{\mathrm{p}-\mathrm{r}}+\sqrt{\mathrm{q}-\mathrm{r}}⩽\min \{\sqrt{\frac{\mathrm{pq}}{\mathrm{r}}},\sqrt{2(\mathrm{p}+\mathrm{q}-2\mathrm{r})}\}$$ $$$$ $$$$

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