Question Number 175593 by Shrinava last updated on 03/Sep/22
$$\mathrm{If}\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}\:\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\begin{vmatrix}{\mathrm{1}}&{\mathrm{a}}&{\mathrm{log}\left(\mathrm{a}^{\boldsymbol{\mathrm{a}}} \right)}\\{\mathrm{1}}&{\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}}\:\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{8}}}\:\centerdot\:\mathrm{log}\left(\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\right)}&{}\\{\mathrm{1}}&{\mathrm{b}}&{\mathrm{log}\left(\mathrm{b}^{\boldsymbol{\mathrm{b}}} \right)}\end{vmatrix}\geqslant\:\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 03/Sep/22
$$\begin{vmatrix}{\mathrm{1}}&{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}}&{\mathrm{log}\left(\mathrm{a}^{\boldsymbol{\mathrm{a}}} \right)}\\{\mathrm{1}}&{\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}}\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{8}}}\:\centerdot\:\mathrm{log}\left(\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\right)}&{\:\:\:\:\:\:?}\\{\mathrm{1}}&{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}}&{\mathrm{log}\left(\mathrm{b}^{\boldsymbol{\mathrm{b}}} \right)}\end{vmatrix}\geqslant\:\mathrm{0} \\ $$$$ \\ $$
Commented by JDamian last updated on 03/Sep/22
$${both}\:{square}\:{roots}\:{lie}\:{on}\:{the} \\ $$$${second}\:{column}? \\ $$
Commented by Shrinava last updated on 03/Sep/22
$$\mathrm{1}\:\:\:\:\:\:\:\:\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}}\:\:\:\:\:\:\:\:\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{8}}}\:\centerdot\:{log}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$