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Question Number 175593 by Shrinava last updated on 03/Sep/22

$$\mathrm{If}0<\mathrm{a}⩽\mathrm{b}\mathrm{then}\mathrm{prove}\mathrm{that}:$$ $$\left|\begin{array}{ccc}1& \mathrm{a}& \log \left({\mathrm{a}}^{\mathbf{a}}\right)\\ 1& \sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}}\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{8}}\cdot \log \left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\right)& \\ 1& \mathrm{b}& \log \left({\mathrm{b}}^{\mathbf{b}}\right)\end{array}\right|⩾0$$

Commented byRasheed.Sindhi last updated on 03/Sep/22

$$\left|\begin{array}{ccc}1& \mathrm{a}& \log \left({\mathrm{a}}^{\mathbf{a}}\right)\\ 1& \sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}}\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{8}}\cdot \log \left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\right)& ?\\ 1& \mathrm{b}& \log \left({\mathrm{b}}^{\mathbf{b}}\right)\end{array}\right|⩾0$$ $$$$

Commented byJDamian last updated on 03/Sep/22

$$bothsquarerootslieonthe$$ $$secondcolumn?$$

Commented byShrinava last updated on 03/Sep/22

$$1\sqrt{\frac{a^2+b^2}{2}}\sqrt{\frac{a^2+b^2}{8}}\cdot log\left(\frac{a^2+b^2}{2}\right)$$

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